EXERCISE 3A

Solve each of the following systems of equations graphically.
Question 1.
2x + 3y = 2,
x – 2y = 8. [CBSE2007]
Solution:
2x + 3y = 2 …..(i)
x – 2y = 8 …(ii)
From Eq. (i),
⇒ 2x = 2 – 3y
⇒x =
Giving some different values to y, we get corresponding values of x as given below:

Now, plot the points (1, 0), (-2, 2) and (4, -2) on the graph and join them to get a line.
Similarly x – 2y = 8 ⇒ x = 8 + 2y

Now plot the points (6, -1), (4, -2) and (2, -3) on the graph and join them to get another line.
We see that these two lines intersect each other at point(4, -2).
x = 4, y = -2

Question 2.
3x + 2y = 4,
2x – 3y = 7. [CBSE2006C]
Solution:
3x + 2y = 4
⇒ 3x = 4 – 2y
⇒ x =
Giving some different values to y, we get corresponding values of x as given below:

Now plot the points (0, 2), (2, -1) and (4, -4) on the graph and join them to get a line. Similarly,
2x – 3y = 7
2x = 3y+ 7
x =

Now plot the points on the graph and join them to get another line.
We see that these two lines intersect each other at point (2, -1).
x = 2, y = -1

Question 3.
2x + 3y = 8,
x – 2y + 3 = 0. [CBSE 2005]
Solution:
2x + 3y = 8
⇒ 2x = 8 – 3y
x =
Now, giving some different values to y, we get corresponding values of x as given below

Now, we plot the points (4,0), (2,3) and (0, 6) on the graph and join them to get a line.
Similarly,
x – 2y + 3 = 0
x = 2y – 3

Now, plot the points (-3, 0), (-1, 1) and (1, 2) on the graph and join them to get another line.
We see that these two lines intersect each other at the point (1, 2).
x = 1, y = 2

Question 4.
2x – 5y + 4 = 0,
2x + y – 8 = 0. [CBSE 2005]
Solution:
2x – 5y + 4 = 0
⇒ 2x = 5y – 4
x =
Giving some different values to y, we get corresponding values of x as given below:

Now, plot the points (-2, 0) (3, 2) and (8,4) on the graph and join them to get a line.
Similarly,
2x + y – 8 = 0
⇒ y = 8 – 2x

Plot the points (1, 6), (2, 1) and (3, 2) on the graph and join them to get another line.
We see that these two lines intersect each other at the point (3, 2).
x = 3, y = 2

Question 5.
3x + 2y = 12,
5x – 2y = 4. [CBSE 2006]
Solution:
3x + 2y = 12
⇒ 3x = 12 – 2y
x =
Giving some different values to y, we get corresponding the values of x as given below:

Now, we plot the points (4,0), (2,3) and (0, 6) on the graph and join them to get a line.
Similarly,
5x – 2y = 4
⇒ 5x = 4 + 2 y
⇒ x =

Plot the points (0, -2), (2, 3) and (4, 8) on the graph and join them to get another line.
We see that these two lines intersect each other at the point (2, 3).
x = 2, y = 3

Question 6.
3x + y + 1 = 0,
2x – 3y + 8 = 0. [CBSE 2007C]
Solution:
3x + y + 1 = 0 ⇒y = -3x – 1
Giving some different values to x, we get corresponding values of y as given below:

Now, plot the points (0, -1), (-1, 2) and (-2, 5) on the graph and join them to get a line.
Similarly,
2x – 3y + 8 = 0
2x = 3y – 8
x =

Now, plot the points (-4, 0), (-1, 2) and (2, 4) on the graph and join them to get another line.
We see that these two lines intersect each other at the points (-1, 2).
x = -1, y = 2

Question 7.
2x + 3y + 5 = 0,
3x – 2y – 12 = 0.
Solution:
2x + 3y + 5 = 0
2x = -3y – 5
x =
Giving some different values to y, we get corresponding values of x as given below:

Now, plot the points (-4, 1), (-1, -1) and (2, -3) on the graph and join them to get a line. Similarly,
3x – 2y – 12 = 0
⇒ 3x = 2y + 12
x =

Plot the points (4, 0), (0, -6) and (2, -3) on the graph and join them to get another line.
We see that these two lines intersect each other at the point (2, -3).
x = 2, y = -3

Question 8.
2x – 3y + 13 = 0,
3x – 2y + 12 = 0.
Solution:
2x – 3y + 13 = 0
⇒ 2x = 3y – 13
⇒ x =
Giving some different values to y, we get corresponding values of x as given below:

Now, plot the points (-5, 1), (-2, 3) and (1, 5) on the graph and join them to get a line.
Similarly,
3x – 2y + 12 = 0
3x = 2y – 12
x =

Now, plot the points (-4, 0), (-2, 3) and (0, 6) on the graph and join them to get another line.
We see that these lines intersect each other at the point (-2, 3).
x = -2, y = 3

Question 9.
2x + 3y – 4 = 0,
3x – y + 5 = 0. [CBSE 2004C]
Solution:
2x + 3y – 4 = 0
⇒ 2x = 4 – 3y
⇒ x =
Giving some different values to y, we get corresponding values of x as given below:

Plot the points (2, 0) (-1, 2) and (5, -2) on the graph and join them to get a line.
Similarly,
3x – y + 5 = 0
⇒ -y = -5 – 3x
⇒ y = 5 + 3x

Plot the points (0, 5), (-1, 2) and (-2, -1) on the graph and join them to get another line.
We see that these two lines intersect each other at the point (-1, 2).
x = -1, y = 2

Question 10.
x + 2y + 2 = 0,
3x + 2y – 2 = 0.
Solution:
x + 2y + 2 = 0
⇒ x = – (2y + 2)
Giving some different values to y, we get corresponding the values of x as given below:

Plot the points (-2,0), (0, -1) and (2, -2) on the graph and join them to get a line.
Similarly,
3x + 2y – 2 = 0
⇒ 3x = 2 – 2y
x =

Plot the points (0, 1), (2, -2) and (-2, 4) on the graph and join them to get another line.
We see that these two lines intersect each other at the point (2, -2).
x = 2, y = -2

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the x-axis:
Question 11.
x – y + 3 = 0, 2x + 3y – 4 = 0.
Solution:
x – y + 3 = 0
⇒ x = y – 3
Giving some different value to y, we get corresponding values of x as given below:

Plot the points (-3, 0), (-1, 2) and (0, 3) on the graph and join them to get a line.
Similarly,
2x + 3y – 4 = 0
⇒ 2x = 4 – 3y
x =

Plot the points (2, 0), (-1, 2) and (5, -2) on the graph and join them to get another line.
We see that these line intersect each other at (-1, 2) and x-axis at A (-3, 0) and D (2, 0).

Area of ∆BAD = x base x altitude
= x AD x BL
= x 5 x 2 = 5 sq.units

Question 12.
2x – 3y + 4 = 0, x + 2y – 5 = 0. [CBSE 2005]
Solution:
2x – 3y + 4 = 0
⇒ 2x = 3y – 4
x =
Giving some different values to y, we get corresponding value of x as given below:

Plot the points (-2, 0), (1, 2) and (4, 4) on the graph and join them to get a line.
Similarly,
x + 2y – 5 = 0
x = 5 – 2y

Now, plot the points (5, 0), (3, 1) and (1, 2) on the graph and join them to get another line.
We see that there two lines intersect each other at the point B (1, 2) and intersect x- axis at A (-2, 0) and D (5, 0) respectively.

Now, area of ∆BAD = x base x altitude
= x AD x BL
= x 7 x 2
= 7 sq. units

Question 13.
4x – 3y + 4 = 0, 4x + 3y – 20 = 0. [CBSE 2008]
Solution:
4x – 3y + 4 = 0
⇒ 4x = 3y – 4
x =
Giving some different values to y, we get corresponding value of x as given below:

Plot the points (-1, 0), (2, 4) and (-4, -4) on the graph and join them, to get a line.
Similarly,
4x + 3y – 20 = 0
4x = 20 – 3y
x =

Plot the points (5, 0), (2, 4) and (-1, 8) on the graph and join them to get a line.

We see that these two lines intersect each other at the point B (2, 4) and intersect x-axis at A (-1, 0) and D (5, 0).
Area ∆BAD = x base x altitude
= x AD x BL
= x 6 x 4 = 12 sq. units

Question 14.
x – y + 1 = 0, 3x + 2y – 12 = 0. [CBSE 2002]
Solution:
x – y + 1 = 0 ⇒ x = y – 1
Giving some different values toy, we get the values of x as given below:

Plot the points (-1, 0), (0, 1) and (1, 2) on the graph and join them to get a line.
Similarly,
3x + 2y – 12 = 0
⇒ 3x = 12 – 2y
x =

Plot the points (4, 0), (2, 3) and (0, 6) on the graph and join them to get another line.

We see that these two lines intersect each ohter at the point E (2, 3) and intersect x- axis at A (-1, 0) and D (4, 0).
Area of ∆EAD = x base x altitude 1
= x AD x EL
= x 5 x 3
=
= 7.5 sq. units

Question 15.
x – 2y + 2 = 0, 2x + y – 6 = 0.
Solution:
x – 2y + 2 = 0 ⇒ x = 2y – 2
Giving some different values to y, we get corresponding values of x as given below:

Plot the points (-2, 0), (0, 1), (2, 2) on the graph and join them to get a line.
Similarly,
2x + y – 6 = 0
⇒ y = 6 – 2x

Plot the points on the graph and join them to get a line.

We see that these two lines intersect each other at the point C (2, 2) and intersect the x-axis at A (-2, 0) and F (3, 0).
Now, area of ∆CAF = x base x altitude
= x AF x CL
= x 5 x 2 = 5 sq. units

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis:
Question 16.
2x – 3y + 6 = 0, 2x + 3y – 18 = 0. [CBSE 2004]
Solution:
2x – 3y + 6 = 0
⇒ 2x = 3y – 6
x =
Giving some different values to y, we get corresponding values of x, as given below:

Now, plot the points (-3, 0), (0, 2), (3, 4) on the graph and join them to get a line.
Similarly,
2x + 3y – 18 = 0
2x = 18 – 3y
x =

Plot the points (0, 6), (3, 4) and (6, 2) on the graph and join them to get a line.
We see that these two lines intersect each other at C (3, 4) and intersect y-axis at B (0, 2) and D (0, 6).

Now, area of ∆CBD = x base x altitude
= x BD x CL
= x 4 x 3 sq. units = 6 sq. units

Question 17.
4x – y – 4 = 0, 3x + 2y – 14 = 0. [CBSE 2006C]
Solution:
4x – y – 4 = 0
⇒ 4x = y + 4
x =
Giving some different values to y, we get corresponding values of x as given below:

Plot the points (1, 0), (0, -4) and (2, 4) on the graph and join them to get a line.
Similarly,
3x + 2y – 14 = 0
2y = 14 – 3x
y =

Now, plot the points (0, 7), (2, 4) and (4, 1) on the graph and join them to get another line.
We see that these two lines intersect each other at C (2, 4) and y-axis at B (0, -4) and D (0, 7) respectively.

Area of ∆CBD = x base x altitude
= x BD x CL
= x 11 x 2 sq. units = 11 sq. units

Question 18.
x – y – 5 = 0, 3x + 5y – 15 = 0 [CBSE 2009C]
Solution:
x – y – 5 = 0 ⇒ x = y + 5
Giving some different values to y, we get corresponding values of x as given below:

Plot the points (5, 0), (0, -5) and (1, -4) on the graph and join these to get a line.
Similarly,
3x + 5y – 15 = 0
3x = 15 – 5y
x =

Plot the points (5, 0), (0, 3) and (-5, 6) on the graph and join them to get a line.
We see that these two lines intersect each other at A (5, 0) and y-axis at B (0, -5) and E (0, 3) respectively.

Now area of ∆ABE = x base x altitude
= x BE x AL
= x 8 x 5 sq. units = 20 sq. units

Question 19.
2x – 5y + 4 = 0, 2x + y – 8 = 0. [CBSE 2005]
Solution:
2x – 5y + 4 = 0
⇒ 2x = 5y – 4
x =
Giving some different values to y, we get corresponding values of x as given below:

Plot the points (-2, 0), (3, 2) and (-7, -2) on the graph and join them to get a line.
Similarly,
2x + y – 8 = 0
2x = 8 – y
x =

Plot the points (4, 0), (3, 2) and (2, 4) on the graph and join them to get a line.
We see that these two lines intersect each other at B (3, 2) and y-axis at G (0, 1) and H (0, 8) respectively.

Now area of ∆EGH = x base x altitude
= x GH x EL
= x 7 x 3 sq. units
= = 10.5 sq. units

Question 20.
5x – y – 7 = 0, x – y + 1 = 0.
Solution:
5x – y – 7 = 0 ⇒ y = 5x – 7
Giving some different values to x, we get corresponding values of y as given below:

Plot the points (0, -7), (1, -2), (2, 3) on the graph and join them to get a line.
Similarly,
x – y + 1 = 0 ⇒ x = y – 1

Plot the points (-1, 0), (1, 2) and (2, 3) on the graph and join them to get a line.

We see that these two lines intersect each other at C (2, 3) and intersect y-axis at A (0, -7) and F (0, 1) respectively.
Now, area of ∆CAF = x base x altitude
= x AF x CL
= x 8 x 2 = 8 sq.units

Question 21.
2x – 3y = 12, x + 3y = 6. [CBSE 2008]
Solution:
2x – 3y – 12 ⇒ 2x = 12 + 3y
x =
Giving some different values to y, we get corresponding values of x as given below:

Plot the points (6, 0), (3, -2) and (0, -4) on the graph and join them to get a line.
Similarly,
x + 3y = 6 ⇒ x = 6 – 3y

Plot the points (6, 0), (0, 2) and (-6, 4) on the graph and join them to get a line.
We see that these two lines intersect each other at the points A (6, 0) and intersect the y-axis at C (0, -4) and E (0, 2) respectively.

Now area of ∆ACE = x base x altitude
= x CE x AO
= x 6 x 6 sq. units = 18 sq.units

Show graphically that each of the following given systems of equations has infinitely many solutions:
Question 22.
2x + 3y = 6, 4x + 6y = 12. [CBSE 2010]
Solution:
2x + 3y = 6
2x = 6 – 3y
x =
Giving some different values to y, we get the corresponding val ues of x as given below:

Plot the points (3, 0), (0, 2) and (-3, 4) on the graph and join them to get a line.
Similarly,
4x + 6y = 12
4x = 12 – 6y
x =

on the graph and join them to get a line.
We see that all the points lie on the same straight line.
This system has infinite many solutions.

Question 23.
3x – y = 5, 6x – 2y = 10.
Solution:
3x – y = 5
⇒ y = 3x – 5
Giving some different values to x, we get corresponding values of y as shown below:

Now, plot the points (0, -5), (1, -2), (2, 1) on the graph and join them to get a line:
Similarly,
6x – 2y = 10
⇒ 6x = 10 + 2y
x =

Plot the points (2, 1), (4, 7), (3, 4) on the graph and join them to get another line.
We see that these lines coincide each other.
This system has infinite many solutions.

Question 24.
2x + y = 6, 6x + 3y = 18.
Solution:
2x + y = 6
y = 6 – 2x
Giving some different values to x, we get corresponding values of y as given below:

Plot the points (0, 6), (2, 2), (4, -2) on the graph and join them to get a line.
Similarly,
6x + 3y = 18
⇒ 6x = 18 – 3y
⇒ x =

Now, plot the points (3, 0), (1, 4) and (5, -4) on the graph and join them to get another line.
We see that these two lines coincide each other.
This system has infinitely many solutions

Question 25.
x – 2y = 5, 3x – 6y = 15.
Solution:
x – 2y = 5 ⇒ x = 5 + 2y
Giving some different values to y, we get corresponding values of x as shown below:

Plot the points (5, 0), (3, -1), (1, -2) on the graph and join them to get a line.
similarly,
3x – 6y = 15
⇒ 3x = 15 + 6y
x =

Plot the points (7, 1), (-1, -3) and (-3, -4) on the graph and join them to get another line.
We see that all the points lie on the same line.
Lines coincide each other.
Hence, the system has infinite many solutions.

Show graphically that each of the following given systems of equations is inconsistent, i.e., has no solution:
Question 26.
x – 2y = 6, 3x – 6y = 0.
Solution:
x – 2y = 6 ⇒ x = 6 + 2y
Giving some different values to y, we get corresponding values of x as given below:

Plot the points (6, 0), (4, -1) and (0, -3) on the graph and join them to get a line.
Similarly,
3x – 6y = 0
⇒ 3x = 6y
⇒ x = 2y

Plot the points (0, 0), (2, 1), (4, 2) on the graph and join them to get a line.

We see that these two lines are parallel i.e., do not intersect each other.
This system has no solution.

Question 27.
2x + 3y = 4, 4x + 6y = 12.
Solution:
2x + 3y = 4
⇒ 2x = 4 – 3y
⇒ x =
Giving some different values to y, we get corresponding values of x as given below:

Plot the points (2, 0), (-1, 2) and (5, -2) on the graph and join them to get a line.
Similarly,
4x + 6y = 12 ⇒ 4x = 12 – 6y
x =

Plot the points (3, 0), (0, 2) and (-3, 4) on the graph and join them to get another line.
We see that two lines are parallel i.e., these do not intersect each other at any point.
Therefore the system has no solution.

Question 28.
2x + y = 6, 6x + 3y = 20.
Solution:
2x + y = 6, y = 6 – 2x
Giving some different values to x, we get corresponding values ofy as given below:

Plot the points (0, 6), (1, 4) and (3, 0) on the graph and join them to get a line.
Similarly,
6x + 3y = 20 ⇒ 6x = 20 – 3y

We see that these two lines are parallel and do not intersect each other.
Therefore this system has no solution.

Question 29.
Draw the graphs of the following equations on the same graph paper:
2x + y = 2, 2x + y = 6.
Find the coordinates of the vertices of the trapezium formed by these lines. Also, find the area of the trapezium so formed. [HOTS]
Solution:
2x + y = 2 ⇒ y = 2 – 2x
Giving some different values to x, we get corresponding values of y as given below:

Plot the points (0, 2), (1, 0) and (-2, 6) on the graph and join them to get a line.
Similarly,
2x + y = 6 ⇒ y = 6 – 2x

Plot the points (0, 6), (2, 2) and (3, 0) on the graph and join them to get another line.
ABFD is the trapezium whose vertices are A (0, 2), B (1, 0), F (3, 0), D (0, 6).

Area of trapezium ABFD = Area ∆DOF – Area ∆AOB
= (DO x OF) – (AO x OB)
= (6 x 3) – (2 x 1) sq.units
= 9 – 1 = 8 sq.units

EXERCISE 3B

Solve for x and y:
Question 1.
x + y = 3,
4x – 3y = 26.
Solution:
x + y = 3 …..(i)
4x – 3y = 26 …(ii)
From (i), x = 3 – y
Substituting the value of x in (ii),
4(3 – y) – 3y = 26
=> 12 – 4y – 3y = 26
-7y = 26 – 12 = 14
y = -2
x = 3 – y = 3 – (-2) = 3 + 2 = 5
Hence, x = 5, y = -2

Question 2.
x – y = 3,
+ = 6
Solution:

Question 3.
2x + 3y = 0,
3x + 4y = 5.
Solution:
2x + 3y= 0 ……..(i)
3x + 4y = 5 …….(ii)

Question 4.
2x – 3y = 13,
7x – 2y = 20.
Solution:
2x – 3y = 13 ……(i)
7x – 2y = 20 ……….(ii)

Question 5.
3x – 5y – 19 = 0,
-7x + 3y + 1 = 0.
Solution:

=> x = -2, y = -5

Question 6.
2x – y + 3 = 0,
3x – 7y + 10 = 0.
Solution:

Question 7.

Solution:

Question 8.


Solution:

Question 9.
4x – 3y = 8,
6x – y =
Solution:

Hence, x = , y =

Question 10.
2x – = 3,
5x = 2y + 7.
Solution:

Question 11.
2x + 5y = ,
3x – 2y = .
Solution:

Question 12.
2x + 3y + 1 = 0,
= y
Solution:

Question 13.
0.4x + 0.3y = 1.7,
0.7x – 0.2y = 0.8.
Solution:
0.4x + 0.3y = 1.7
0.7x – 0.2y = 0.8
Multiplying each term by 10
4x + 3y = 17
0.7x – 2y = 8
Multiply (i) by 2 and (ii) by 3,
8x + 6y = 34
21x – 6y = 24
Adding, we get
29x = 58
x = 2
From (i) 4 x 2 + 3y = 17
=> 8 + 3y = 17
=> 3y = 17 – 8 = 9
y = 3
x = 2, y = 3

Question 14.
0.3x + 0.5y = 0.5,
0.5x + 0.7y = 0.74.
Solution:

Question 15.
7(y + 3) – 2(x + 2) = 14,
4(y – 2) + 3(x – 3) = 2.
Solution:

Question 16.
6x + 5y = 7x + 3y + 1 = 2(x + 6y – 1)
Solution:
6x + 5y = 7x + 3y + 1 = 2(x + 6y – 1)
6x + 5y = 7x + 3y + 1
=> 6x + 5y – 7x – 3y = 1
=> -x + 2y = 1
=> 2y – x = 1 …(i)
7x + 3y + 1 = 2(x + 6y – 1)
7x + 3y + 1 = 2x + 12y – 2
=> 7x + 3y – 2x – 12y = -2 – 1
=> 5x – 9y = -3 …..(ii)
From (i), x = 2y – 1
Substituting the value of x in (ii),
5(2y – 1) – 9y = -3
=> 10y – 5 – 9y = -3
=> y = -3 + 5
=> y = +2
x = 2y – 1 = 2 x 2 – 1 = 4 – 1 = 3
x = 3, y = 2

Question 17.

Solution:

and x = y – 4 = 6 – 4 = 2
x = 2, y = 6

Question 18.
+ 6y = 13,
+ 4y = 7 (x ≠ 0)
Solution:

Question 19.
x + = 6,
3x – = 5 (y ≠ 0)
Solution:

Question 20.
2x – = 9,
3x + = 2 (y ≠ 0)
Solution:

x = 3, y = -1

Question 21.

Solution:

Question 22.

Solution:

Question 23.

Solution:

Question 24.

Solution:

Question 25.
4x + 6y = 3xy,
8x + 9y = 5xy (x ≠ 0, y ≠ 0).
Solution:

Question 26.
x + y = 5xy,
3x + 2y = 13xy (x ≠ 0, y ≠ 0).
Solution:

Question 27.

Solution:


x = 3, y = 2

Question 28.

Solution:

Question 29.

Solution:

Question 30.

Solution:

Question 31.

Solution:

Question 32.
71x + 37y = 253,
37x + 71y = 287. [CBSE 2007C]
Solution:
71x + 37y = 253
37x + 71y = 287
Adding, we get
108x + 108y = 540
x + y = 5 ………(i) (Dividing by 108)
and subtracting,
34x – 34y = -34
x – y = -1 ……..(ii) (Dividing by 34)
Adding, (i) and (ii)
2x = 4 => x = 2
and subtracting,
2y = 6 => y = 3
Hence, x = 2, y = 3

Question 33.
217x+ 131y = 913,
131x + 217y = 827.
Solution:
217x + 131y = 913 …(i)
131x + 217y = 827 …..(ii)
Adding, we get
348x + 348y = 1740
x + 7 = 5 …..(iii) (Dividing by 348)
and subtracting,
86x – 86y = 86
x – y = 1 …(iv) (Dividing by 86)
Now, adding (iii) and (iv)
2x = 6 => x = 3
and subtracting,
2y = 4 => y = 2
x = 3, y = 2

Question 34.
23x – 29y = 98,
29x – 23y = 110.
Solution:
23x – 29y = 98 ……(i)
29x – 23y = 110 ……(ii)
Adding, we get
52x – 52y = 208
x – y = 4 ……(iii) (Dividing by 52)
and subtracting
-6x – 6y = -12
x + y = 2 …..(iv)
Adding (iii), (iv)
2x = 6 => x = 3
Subtracting (iii) from (iv)
2y = -2 => y = -1
x = 3, y = -1

Question 35.

Solution:

x = 1 and y = 2

Question 36.

Solution:

Question 37.

Solution:

Question 38.

Solution:

Question 39.
3 (2x +y) = 7xy,
3 (x + 3y) = 11xy (x ≠ 0 and y ≠ 0).
Solution:

Question 40.
x + y = a + b,
ax – by = a² – b².
Solution:

Question 41.

Solution:

Question 42.
px + qy = p – q,
qx – py = p + q.
Solution:

Question 43.

Solution:

Question 44.
6 (ax + by) = 3a + 2b,
6 (bx – ay) = 3b – 2a.
Solution:


a = , b =

Question 45.
ax – by = a² + b²,
x + y = 2a. [CBSE 2006C]
Solution:

Question 46.

Solution:

Question 47.

Solution:

Question 48.
x + y = a + b,
ax – by = a² – b².
Solution:

Question 49.
a²x + b²y = c²,
b²x + a²y = d².
Solution:
a²x + b²y = c² ……(i)
b²x + a²y = d² …….(ii)
Multiply (i) by a² and (ii) by b²,

Question 50.

Solution:

EXERCISE 3C

Solve for x and y:
Question 1.
x + y = 3,
4x – 3y = 26.
Solution:
x + y = 3 …..(i)
4x – 3y = 26 …(ii)
From (i), x = 3 – y
Substituting the value of x in (ii),
4(3 – y) – 3y = 26
=> 12 – 4y – 3y = 26
-7y = 26 – 12 = 14
y = -2
x = 3 – y = 3 – (-2) = 3 + 2 = 5
Hence, x = 5, y = -2

Question 2.
x – y = 3,
+ = 6
Solution:

Question 3.
2x + 3y = 0,
3x + 4y = 5.
Solution:
2x + 3y= 0 ……..(i)
3x + 4y = 5 …….(ii)

Question 4.
2x – 3y = 13,
7x – 2y = 20.
Solution:
2x – 3y = 13 ……(i)
7x – 2y = 20 ……….(ii)

Question 5.
3x – 5y – 19 = 0,
-7x + 3y + 1 = 0.
Solution:

=> x = -2, y = -5

Question 6.
2x – y + 3 = 0,
3x – 7y + 10 = 0.
Solution:

Question 7.

Solution:

Question 8.


Solution:

Question 9.
4x – 3y = 8,
6x – y =
Solution:

Hence, x = , y =

Question 10.
2x – = 3,
5x = 2y + 7.
Solution:

Question 11.
2x + 5y = ,
3x – 2y = .
Solution:

Question 12.
2x + 3y + 1 = 0,
= y
Solution:

Question 13.
0.4x + 0.3y = 1.7,
0.7x – 0.2y = 0.8.
Solution:
0.4x + 0.3y = 1.7
0.7x – 0.2y = 0.8
Multiplying each term by 10
4x + 3y = 17
0.7x – 2y = 8
Multiply (i) by 2 and (ii) by 3,
8x + 6y = 34
21x – 6y = 24
Adding, we get
29x = 58
x = 2
From (i) 4 x 2 + 3y = 17
=> 8 + 3y = 17
=> 3y = 17 – 8 = 9
y = 3
x = 2, y = 3

Question 14.
0.3x + 0.5y = 0.5,
0.5x + 0.7y = 0.74.
Solution:

Question 15.
7(y + 3) – 2(x + 2) = 14,
4(y – 2) + 3(x – 3) = 2.
Solution:

Question 16.
6x + 5y = 7x + 3y + 1 = 2(x + 6y – 1)
Solution:
6x + 5y = 7x + 3y + 1 = 2(x + 6y – 1)
6x + 5y = 7x + 3y + 1
=> 6x + 5y – 7x – 3y = 1
=> -x + 2y = 1
=> 2y – x = 1 …(i)
7x + 3y + 1 = 2(x + 6y – 1)
7x + 3y + 1 = 2x + 12y – 2
=> 7x + 3y – 2x – 12y = -2 – 1
=> 5x – 9y = -3 …..(ii)
From (i), x = 2y – 1
Substituting the value of x in (ii),
5(2y – 1) – 9y = -3
=> 10y – 5 – 9y = -3
=> y = -3 + 5
=> y = +2
x = 2y – 1 = 2 x 2 – 1 = 4 – 1 = 3
x = 3, y = 2

Question 17.

Solution:

and x = y – 4 = 6 – 4 = 2
x = 2, y = 6

Question 18.
+ 6y = 13,
+ 4y = 7 (x ≠ 0)
Solution:

Question 19.
x + = 6,
3x – = 5 (y ≠ 0)
Solution:

Question 20.
2x – = 9,
3x + = 2 (y ≠ 0)
Solution:

x = 3, y = -1

Question 21.

Solution:

Question 22.

Solution:

Question 23.

Solution:

Question 24.

Solution:

Question 25.
4x + 6y = 3xy,
8x + 9y = 5xy (x ≠ 0, y ≠ 0).
Solution:

Question 26.
x + y = 5xy,
3x + 2y = 13xy (x ≠ 0, y ≠ 0).
Solution:

Question 27.

Solution:


x = 3, y = 2

Question 28.

Solution:

Question 29.

Solution:

Question 30.

Solution:

Question 31.

Solution:

Question 32.
71x + 37y = 253,
37x + 71y = 287. [CBSE 2007C]
Solution:
71x + 37y = 253
37x + 71y = 287
Adding, we get
108x + 108y = 540
x + y = 5 ………(i) (Dividing by 108)
and subtracting,
34x – 34y = -34
x – y = -1 ……..(ii) (Dividing by 34)
Adding, (i) and (ii)
2x = 4 => x = 2
and subtracting,
2y = 6 => y = 3
Hence, x = 2, y = 3

Question 33.
217x+ 131y = 913,
131x + 217y = 827.
Solution:
217x + 131y = 913 …(i)
131x + 217y = 827 …..(ii)
Adding, we get
348x + 348y = 1740
x + 7 = 5 …..(iii) (Dividing by 348)
and subtracting,
86x – 86y = 86
x – y = 1 …(iv) (Dividing by 86)
Now, adding (iii) and (iv)
2x = 6 => x = 3
and subtracting,
2y = 4 => y = 2
x = 3, y = 2

Question 34.
23x – 29y = 98,
29x – 23y = 110.
Solution:
23x – 29y = 98 ……(i)
29x – 23y = 110 ……(ii)
Adding, we get
52x – 52y = 208
x – y = 4 ……(iii) (Dividing by 52)
and subtracting
-6x – 6y = -12
x + y = 2 …..(iv)
Adding (iii), (iv)
2x = 6 => x = 3
Subtracting (iii) from (iv)
2y = -2 => y = -1
x = 3, y = -1

Question 35.

Solution:

x = 1 and y = 2

Question 36.

Solution:

Question 37.

Solution:

Question 38.

Solution:

Question 39.
3 (2x +y) = 7xy,
3 (x + 3y) = 11xy (x ≠ 0 and y ≠ 0).
Solution:

Question 40.
x + y = a + b,
ax – by = a² – b².
Solution:

Question 41.

Solution:

Question 42.
px + qy = p – q,
qx – py = p + q.
Solution:

Question 43.

Solution:

Question 44.
6 (ax + by) = 3a + 2b,
6 (bx – ay) = 3b – 2a.
Solution:


a = , b =

Question 45.
ax – by = a² + b²,
x + y = 2a. [CBSE 2006C]
Solution:

Question 46.

Solution:

Question 47.

Solution:

Question 48.
x + y = a + b,
ax – by = a² – b².
Solution:

Question 49.
a²x + b²y = c²,
b²x + a²y = d².
Solution:
a²x + b²y = c² ……(i)
b²x + a²y = d² …….(ii)
Multiply (i) by a² and (ii) by b²,

Question 50.

Solution:

EXERCISE 3D

Show that each of the following systems of equations has a unique solution and solve it:
Question 1.
3x + 5y = 12, 5x + 3y = 4
Solution:

Question 2.
2x – 3y = 17, 4x + y = 13.
Solution:

Question 3.
+ = 3, x – 2y = 2
Solution:

This system has a unique solution.
From (ii), x = 2 + 2y
Substituting the value of x in (i),
2(2 + 2y) + 3y = 18
=> 4 + 4y + 3y = 18
=> 7y = 18 – 4 = 14
=> y = 2
and x = 2 + 2 x 2 = 2 + 4 = 6
x = 6, y = 2

Find the value of k for which each of the following systems of equations has a unique solution:
Question 4.
2x + 3y – 5 = 0, kx – 6y – 8 = 0.
Solution:

Question 5.
x – ky = 2, 3x + 2y + 5 = 0.
Solution:

Question 6.
5x – 7y – 5 = 0, 2x + ky – 1 = 0.
Solution:

Question 7.
4x + ky + 8 = 0, x + y + 1 = 0.
Solution:

Question 8.
4x – 5y = k, 2x – 3y = 12.
Solution:

Question 9.
kx + 3y = (k – 3), 12x + ky = k.
Solution:

Question 10.
Show that the system of equations 2x – 3y = 5, 6x – 9y = 15 has an infinite number of solutions.
Solution:

Question 11.
Show that the system of equations 6x + 5y = 11, 9x + y = 21 has no solution.
Solution:

Question 12.
For what value of k does the system of equations kx + 2y = 5, 3x – 4y = 10 have
(i) a unique solution,
(ii) no solution?
Solution:

Question 13.
For what value of k does the system of equations x + 2y = 5, 3x + ky + 15 = 0 have
(i) a unique solution,
(ii) no solution?
Solution:

Question 14.
For what value of k does the system of equations x + 2y = 3, 5x + ky + 7 = 0 have
(i) a unique solution,
(ii) no solution? Also, show that there is no value of k for which the given system of equations has infinitely many solutions.
Solution:

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:
Question 15.
2x + 3y = 7,
(k – 1) x + (k + 2) y = 3k. [CBSE 2010]
Solution:

Question 16.
2x + (k – 2 ) y = k,
6x + (2k – 1) y = (2k + 5). [CBSE 2000C]
Solution:

Question 17.
kx + 3y = (2k+ 1),
2(k + 1) x + 9y = (7k + 1). [CBSE 2000C]
Solution:

Question 18.
5x + 2y = 2k,
2(k + 1) x + ky = (3k + 4). [CBSE 2003C]
Solution:

Question 19.
(k – 1) x – y = 5
(k + 1) x + (1 – k) y = (3k + 1) [CBSE 2003]
Solution:

Question 20.
(k – 3) x + 3y = k,
kx + ky = 12.
Solution:

=> k (k – 6) = 0
Either k = 0, which is not true, or k – 6 = 0, then k = 6
k = 6

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:
Question 21.
(a – 1) x + 3y = 2,
6x + (1 – 2b) y = 6. [CBSE 2002C]
Solution:

Question 22.
(2a – 1) x + 3y = 5,
3x + (6 – 1) y = 2. [CBSE 2001C]
Solution:

Question 23.
2x – 3y = 7,
(a + b) x – (a + b – 3) y = 4a + b. [CBSE 2002]
Solution:

Question 24.
2x + 3y = 7,
(a + b + 1) x + (a + 2b + 2) y – 4(a + b) + 1. [CBSE 2003]
Solution:

Question 25.
2x + 3y = 7,
(a + b) x + (2a – b) y = 21. [CBSE 2001]
Solution:

Question 26.
2x + 3y = 7,
2ax + (a + b) y = 28. [CBSE 2001]
Solution:

Find the value of k for which each of the following systems of equations has no solution:
Question 27.
8x + 5y = 9, kx + 10y = 15.
Solution:

Question 28.
kx + 3y = 3, 12x + ky = 6.
Solution:
kx + 3y = 3
12x + ky = 6

Question 29.
3x – y – 5 = 0, 6x – 2y + k = 0 (k ≠ 0). [CBSE 2008]
Solution:

Question 30.
kx + 3y = k – 3, 12x + ky = k. [CBSE 2009]
Solution:

Question 31.
Find the value of k for which the system of equations 5x – 3y = 0, 2x + ky = 0 has a nonzero solution.
Solution:

EXERCISE 3E

Question 1.
5 chairs and 4 tables together cost ₹ 5600, while 4 chairs and 3 tables together cost ₹ 4340. Find the cost of a chair and that of a table.
Solution:
Let cost of one chair = ₹ x
and cost of one table = ₹ y
According to the conditions,
5x + 4y = ₹ 5600 …(i)
4x + 3y = ₹ 4340 …(ii)

x = -560
and from (i)
5 x 560 + 4y = 5600
2800 + 4y = 5600
⇒ 4y = 5600 – 2800
⇒ 4y = 2800
⇒ y = 700
Cost of one chair = ₹ 560
and cost of one table = ₹ 700

Question 2.
23 spoons and 17 forks together cost ₹ 1770, while 17 spoons and 23 forks together cost ₹ 1830. Find the cost of a spoon and that of a fork.
Solution:
Let the cost of one spoon = ₹ x and cost of one fork = ₹ y
According to the conditions,
23x + 17y = 1770 …(i)
17x + 23y = 1830 …(ii)
Adding, we get
40x + 40y = 3600
Dividing by 40,
x + y = 90 …(iii)
and subtracting,
6x – 6y = -60
Dividing by 6,
x – y = -10 …(iv)
Adding (iii) and (iv)
2x = 80 ⇒ x = 40
and subtracting,
2y = 100 ⇒ y = 50
Cost of one spoon = ₹ 40
and cost of one fork = ₹ 50

Question 3.
A lady has only 25-paisa and 50-paisa coins in her purse. If she has 50 coins in all totalling ₹ 19.50, how many coins of each kind does she have?
Solution:
Let number of 25-paisa coins = x
and number 50-paisa coins = y
Total number of coins = 50
and total amount = ₹ 19.50 = 1950 paisa
x + y = 50 …(i)
25x + 50y = 1950
⇒ x + 2y = 78 …(ii)
Subtracting (i) from (ii), y = 28
x = 50 – y = 50 – 28 = 22
Number of 25-paisa coins = 22
and 50-paisa coins = 28

Question 4.
The sum of two numbers is 137 and their difference is 43. Find the numbers.
Solution:
Sum of two numbers = 137
and difference = 43
Let first number = x
and second number = y
x + y = 137 …..(i)
x – y = 43 ……(ii)
Adding, we get
2x = 180 ⇒ x = 90
and subtracting,
2y = 94
y = 47
First number = 90
and second number = 47

Question 5.
Find two numbers such that the sum of twice the first and thrice the second is 92, and four times the first exceeds seven times the second by 2.
Solution:
Let first number = x
and second number = y
According to the conditions,
2x + 3y = 92 …(i)
4x – 7y = 2 …(ii)
Multiply (i) by 2 and (ii) by 1
4x + 6y = 184 …..(iii)
4x – 7y = 2 …….(iv)
Subtracting (iii) from (iv),
13y = 182
y = 14
From (i), 2x + 3y = 92
2x + 3 x 14 = 92
⇒ 2x + 42 = 92
⇒ 2x = 92 – 42 = 50
⇒ x = 25
First number = 25
Second number = 14

Question 6.
Find two numbers such that the sum of thrice the first and the second is 142, and four times the first exceeds the second by 138.
Solution:
Let first number = x
and second number = y
According to the conditions,
3x + y=142 …(i)
4x – y = 138 …(ii)
Adding, we get
7x = 280
⇒ x = 40
and from (i)
3 x 40 + y = 142
⇒ 120 + y = 142
⇒ y = 142 – 120 = 22
First number = 40,
second number = 22

Question 7.
If 45 is subtracted from twice the greater of two numbers, it results in the other number. If 21 is subtracted from twice the smaller number, it results in the greater number. Find the numbers.
Solution:
Let first greater number = x
and second smaller number = y
According to the conditions,
2x – 45 = y …(i)
2y – 21 = x …(ii)
Substituting the value of y in (ii),
2 (2x – 45) – 21 = x
⇒ 4x – 90 – 21 = x
⇒ 4x – x = 111
⇒ 3x = 111
⇒ x = 37
From (i),
y = 2 x 37 – 45 = 74 – 45 = 29
The numbers are 37, 29

Question 8.
If three times the larger of two numbers is divided by the smaller, we get 4 as the quotient and 8 as the remainder. If five times the smaller is divided by the larger, we get 3 as the quotient and 5 as the remainder. Find the numbers.
Solution:
Let larger number = x
and smaller number = y
According to the conditions,
3x = 4 x y + 8 ⇒ 3x = 4y + 8 …….(i)
5y = x x 3 + 5 ⇒ 5y = 3x + 5 …(ii)
Substitute the value of 3x in (ii),
5y = 4y + 8 + 5
⇒ 5y – 4y = 13
⇒ y = 13
and 3x = 4 x 13 + 8 = 60
⇒ x = 20
Larger number = 20
and smaller number = 13

Question 9.
If 2 is added to each of two given numbers, their ratio becomes 1 : 2. However, if 4 is subtracted from each of the given numbers, the ratio becomes 5 : 11. Find the numbers.
Solution:
Let first number = x and
second number = y
According to the conditions,

⇒ 11x – 44 = 5(2x + 2) – 20
⇒ 11x – 44 = 10x + 10 – 20
⇒ 11x – 10x = 10 – 20 + 44
⇒ x = 34
and y = 2 x 34 + 2 = 68 + 2 = 70
Numbers are 34 and 70

Question 10.
The difference between two numbers is 14 and the difference between their squares is 448. Find the numbers.
Solution:
Let first number = x
and second number (smaller) = y
According to the conditions,
x – y = 14
and x² – y² = 448
⇒ (x + y) (x – y) = 448
⇒ (x + y) x 14 = 448
⇒ x + y = 32 ……(i)
and x – y = 14 ……(ii)
Adding (i) and (ii),
2x = 46 ⇒ x = 23
and subtracting (i) and (ii),
2y = 18 ⇒ y = 9
Numbers are 23, 9

Question 11.
The sum of the digits of a two-digit number is 12. The number obtained by interchanging its digits exceeds the given number by 18. Find the number. [CBSE 2006]
Solution:
Let ones digit of a two digit number = x
and tens digit = y
Number = x + 10y
By interchanging the digits,
Ones digit = y
and tens digit = x
Number = y + 10x
According to the conditions,
x + y = 12 ………. (i)
y + 10x = x + 10y + 18
⇒ y + 10x – x – 10y = 18
⇒ 9x – 9y = 18
⇒ x – y = 2 …(ii) (Dividing by 9)
Adding (i) and (ii),
2x = 14 ⇒ x = 7
and subtracting,
2y = 10 ⇒ y = 5
Number = 7 + 10 x 5 = 7 + 50 = 57

Question 12.
A number consisting of two digits is seven times the sum of its digits. When 27 is subtracted from the number, the digits are reversed. Find the number.
Solution:
Let one’s digit of a two digit number = x
and ten’s digit = y
Then number = x + 10y
After reversing the digits,
Ones digit = y
and ten’s digit = x
and number = y + 10x
According to the conditions,
x + 10y – 27 = y + 10x
⇒ y + 10x – x – 10y = -27
⇒ 9x – 9y = -27
⇒ x – y = -3 …(i)
and 7 (x + y) = x + 10y
7x + 7y = x+ 10y
⇒ 7x – x = 10y – 7y
⇒ 6x = 3y
⇒ 2x = y …(ii)
Substituting the value of y in (i)
x – 2x = -3
⇒ -x = -3
⇒ x = 3
y = 2x = 2 x 3 = 6
Number = x + 10y = 3 + 10 x 6 = 3 + 60 = 63

Question 13.
The sum of the digits of a two-digit number is 15. The number obtained by interchanging the digits exceeds the given number by 9. Find the number. [CBSE 2004]
Solution:
Let one’s digit of a two digit number = x
and ten’s digit = y
Then number = x + 10y
After interchanging the digits,
One’s digit = y
and ten’s digit = x
Then number = y + 10x
According to the conditions,
y + 10x = x + 10y + 9
⇒ y + 10x – x – 10y = 9
⇒ 9x – 9y = 9
⇒ x – y = 1 …(i)
and x + y= 15 …(ii)
Adding, we get
2x = 16
x = 8
and subtracting,
2y = 14
⇒ y = 7
Number = x + 10y = 8 + 10 x 7 = 8 + 70 = 78

Question 14.
A two-digit number is 3 more than 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.
Solution:
Let one’s digit of the two digit number = x
and ten’s digit = y
Then number = x + 10y
By reversing the digits,
One’s digit = y
and ten’s digit = x
Then number = y + 10x
Now, according to the conditions,
x + 10y + 18 = y + 10x
⇒ 18 = y + 10x – x – 10y
⇒ 9x – 9y = 18
⇒ x – y = 2 …(i)
and 4(x + y) + 3 = x + 10y
4x + 4y + 3 = x + 10y
⇒ 4x + 4y – x – 10y = -3
3x – 6y = -3
⇒ x – 2y = -1 ……..(ii)
Subtracting,
y = 3
and x = 2y – 1 = 2 x 3 – 1 = 6 – 1 = 5
Number = x + 10y = 5 + 10 x 3 = 5 + 30 = 35

Question 15.
A number consists of two digits. When it is divided by the sum of its digits, the quotient is 6 with no remainder. When the number is diminished by 9, the digits are reversed. Find the number. [CBSE 1999C]
Solution:
Let ones digit of a two digit number = x
and tens digit = y
Then number = x + 10y
By reversing the digits,
One’s digit = y
and ten’s digit = x
and number = y + 10x
According to the conditions,
x + 10y – 9 = y + 10x
⇒ x + 10y – y – 10x = 9
⇒ -9x + 9y = 9
⇒x – y = -1 …(i) (Dividing by -9)

Question 16.
A two-digit number is such that the product of its digits is 35. If 18 is added to the number, the digits interchange their places. Find the number. [CBSE 2006]
Solution:
Let the one’s digit of a two digit number = x
and ten’s digit = y
Then number = x + 10y
By interchanging the digits,
One’s digit = y
and ten’s digit = x
Then number = y + 10x
According to the conditions,
x + 10y + 18 = y + 10x
⇒ 18 = y + 10x – x – 10y
⇒ 9x – 9y = 18
⇒ x – y = 2 …(i)
and xy = 35 …(ii)
Now, (x + y)² = (x – y)² + 4xy = (2)² + 4 x 35 = 4 + 140 = 144 = (12)²
⇒ (x + y) = 12 …(iii)
Subtracting (i) from (iii), we get

Question 17.
A two-digit number is such that the product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number. [CBSE 2006C]
Solution:
Let one’s digit of a two digit number = x
and ten’s digit = y
Then number = x + 10y
After interchanging the digits One’s digit = y
Ten’s digit = x
Then number = y + 10x
According to the conditions,
x + 10y – 63 = y + 10x

Question 18.
The sum of a two-digit number and the number obtained by reversing the order of its digits is 121, and the two digits differ by 3. Find the number.
Solution:
Let one’s digit of a two digit number = x
and ten’s digit = y
Number = x + 10y
By reversing the digits,
One’s digit = y
and ten’s digit = x
Number = y + 10x
According to the conditions,
x + 10y + y + 10x = 121
⇒ 11x + 11y = 121
⇒ x + y = 11 …(i)
x – y = 3 …(ii)
Adding, we get
2x = 14 ⇒ x = 7
Subtracting,
2y = 8 ⇒ y = 4
Number = 7 + 10 x 4 = 7 + 40 = 47
or 4 + 10 x 7 = 4 + 70 = 74

Question 19.
The sum of the numerator and denominator of a fraction is 8. If 3 is added to both of the numerator and the denominator, the fraction becomes . Find the fraction. [CBSE 2003]
Solution:
Let numerator of a fraction = x
and denominator = y
Then fraction =
According to the conditions,
x + y = 8 …(i)

Question 20.
If 2 is added to the numerator of a fraction, it reduces to and if 1 is subtracted from the denominator, it reduces to . Find the fraction.
Solution:
Let numerator of a fraction = x
and denominator = y
Then fraction =
According to the conditions,
=
=
⇒ 2x + 4 = y …(i)
3x = y – 1 …(ii)
⇒ 3x = 2x + 4 – 1
⇒ 3x = 2x + 3
⇒ 3x – 2x = 3
⇒ x = 3
and y = 2x + 4 = 2 x 3 + 4 = 6 + 4 = 10
Fraction =

Question 21.
The denominator of a fraction is greater than its numerator by 11. If 8 is added to both its numerator and denominator, it becomes . Find the fraction.
Solution:
Let numerator of a fraction = x
and denominator = y
Then fraction =
According to the conditions,
y – x = 11
y = 11 + x …(i)

Question 22.
Find a fraction which becomes when 1 is subtracted from the numerator and 2 is added to the denominator, and the fraction becomes when 7 is subtracted from the numerator and 2 is subtracted from the denominator.
Solution:
Let numerator of a fraction = x
and denominator = y
Then fraction =
According to the conditions,

Question 23.
The sum of the numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator are increased by 3, they are in the ratio 2 : 3. Determine the fraction. [CBSE 2010]
Solution:
Let numerator of a fraction = x
and denominator = y
Then fraction =
According to the conditions,
x + y = 4 + 2x
⇒ y = 4 + x …(i)

Fraction = =

Question 24.
The sum of two numbers is 16 and sum of their reciprocal is . Find the numbers. [CBSE 2005]
Solution:
Let first number = x
and second number = y
According to the conditions,
x + y = 16

Question 25.
There are two classrooms A and B. If 10 students are sent from A to B, the number of students in each room becomes the same. If 20 students are sent from B to A, the number of students in A becomes double the number of students in B’. Find the number of students in each room.
Solution:
Let in classroom A, the number of students = x
and in classroom B = y
According to the conditions,
x – 10 = y + 10
⇒ x – y = 10 + 10 = 20
⇒ x – y = 20 …(i)
and x + 20 = 2 (y – 20)
⇒ x + 20 = 2y – 40
⇒ x – 2y = -(40 + 20) = -60
x – 2y = -60 …(ii)
Subtracting, y = 80
and x – y = 20
⇒ x – 80 = 20
⇒ x = 20 + 80 = 100
Number of students in classroom A = 100 and in B = 80

Question 26.
Taxi charges in a city consist of fixed charges and the remaining depending upon the distance travelled in kilometres. If a man travels 80 km, he pays ₹ 1330, and travelling 90 km, he pays ₹ 1490. Find the fixed charges and rate per km.
Solution:
Let fixed charges = ₹ x
and other charges = ₹ y per km
According to the conditions,
For 80 km,
x + 80y = ₹ 1330 …(i)
and x + 90y = ₹ 1490 …(ii)
Subtracting (i) from (ii),
10y = 160 ⇒ y = 16
and from (i)
x + 80 x 16 = 1330
⇒ x + 1280 = 1330
⇒ x = 1330 – 1280 = 50
Fixed charges = ₹ 50
and rate per km = ₹ 16

Question 27.
A part of monthly hostel charges in a college hostel are fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 25 days, he has to pay ₹ 4500, whereas a student B who takes food for 30 days, has to pay ₹ 5200. Find the fixed charges per month and the cost of the food per day.
Solution:
Let fixed charges of the hostel = ₹ x
and other charges per day = ₹ y
According to the conditions,
x + 25y = 4500 ……..(i)
x + 30y = 5200 ……(ii)
Subtracting (i) from (ii),
5y = 700
y = 140
and from (i),
x + 25 x 140 = 4500
⇒ x + 3500 = 4500
⇒ x = 4500 – 3500 = 1000
Fixed charges = ₹ 1000
and per day charges = ₹ 140

Question 28.
A man invested an amount at 10% per annum and another amount at 8% per annum simple interest. Thus, he received ₹ 1350 as annual interest. Had he interchanged the amounts invested, he would have received ₹ 45 less as interest. What amounts did he invest at different rates?
Solution:
Let first investment = ₹ x
and second investment = ₹ y
Rate of interest = 10% p.a. for first kind and 8% per second
Interest is for the first investment = ₹ 1350
and for the second = ₹ 1350 – ₹45 = ₹ 1305
According to the conditions,

Question 29.
The monthly incomes of A and B are in the ratio 5 : 4 and their monthly expenditures are in the ratio 7 : 5. If each saves ₹ 9000 per month, find the monthly income of each.
Solution:
Ratio in the income of A and B = 5 : 4
Let A’s income = ₹ 5x and
B’s income = ₹ 4x
and ratio in their expenditures = 7 : 5
Let A’s expenditure = 7y
and B’s expenditure = 5y
According to the conditions,
5x – 7y = 9000 …(i)
and 4x – 5y = 9000 …(ii)
Multiply (i) by 5 and (ii) by 7,
25x – 35y = 45000
28x – 37y = 63000
Subtracting, we get
3x = 18000
⇒ x = 6000
A’s income = 5x = 5 x 6000 = ₹ 30000
and B’s income = 4x = 4 x 6000 = ₹ 24000

Question 30.
A man sold a chair and a table together for ₹ 1520, thereby making a profit of 25% on chair and 10% on table. By selling them together for ₹ 1535, he would have made a profit of 10% on the chair and 25% on the table. Find the cost price of each.
Solution:
Let cost of one chair = ₹ x
and cost of one table = ₹ y
In first case,
Profit on chair = 25%
and on table = 10%
and selling price = ₹ 1520
According to the conditions,

Question 31.
Points A and B are 70 km apart on a highway. A car starts from A and another car starts from B simultaneously. If they travel in the same direction, they meet in 7 hours. But, if they travel towards each other, they meet in 1 hour. Find the speed of each car. [CBSE 2007C]
Solution:
Distance between two stations A and B = 70 km
Let speed of first car (starting from A) = x km/hr
and speed of second car = y km/hr

According to the conditions,
7x – 7y = 70
⇒ x – y = 10 …(i)
and x + y = 70 …(ii)
Adding (i) and (ii),
2x = 80 ⇒ x = 40
Subtracting (i) and (ii),
2y = 60 ⇒ y = 30
Speed of car A = 40 km/hr
and speed of car B = 30 km/hr

Question 32.
A train covered a certain distance at a uniform speed. If the train had been 5 kmph faster, it would have taken 3 hours less than the scheduled time. And, if the train were slower by 4 kmph, it would have taken 3 hours more than the scheduled time. Find the length of the journey.
Solution:
Let uniform speed of the train = x km/hr
and time taken = y hours
Distance = x x y = xy km
Case I:
Speed = (x + 5) km/hr
and Time = (y – 3) hours
Distance = (x + 5) (y – 3)
(x + 5) (y – 3) = xy
⇒ xy – 3x + 5y – 15 = xy
-3x + 5y = 15 …(i)
Case II:
Speed = (x – 4) km/hr
and Time = (y + 3) hours
Distance = (x – 4) (y + 3)
(x – 4) (y + 3) = xy
⇒ xy + 3x – 4y – 12 = xy
3x – 4y = 12 …(ii)
Adding (i) and (ii),
y = 27
and from (i),
-3x + 5 x 27 = 15
⇒ -3x + 135 = 15
⇒ -3x = 15 – 135 = -120
⇒ x = 40
Speed of the train = 40 km/hr
and distance = 27 x 40 = 1080 km

Question 33.
Abdul travelled 300 km by train and 200 km by taxi taking 5 hours 30 minutes. But, if he travels 260 km by train and 240 km by taxi, he takes 6 minutes longer. Find the speed of the train and that of the taxi. [CBSE 2006C]
Solution:
Let the speed of the train = x km/hr
and speed of taxi = y km/hr
According to the conditions,

Question 34.
Places A and B are 160 km apart on a highway. One car starts from A and another from B at the same time. If they travel in the same direction, they meet in 8 hours. But, if they travel towards each other, they meet in 2 hours. Find the speed of each car. [CBSE 2009C]
Solution:
Distance between stations A and B = 160 km
Let the speed of the car starts from A = x km/hr
and speed of car starts from B = y km/hr
8x – 8y = 160
⇒ x – y = 20 …(i)
and 2x + 2y = 160
⇒ x + y = 80 …(ii)
Adding (i) and (ii)
2x = 100 ⇒ x = 50
and subtracting,
2y = 60 ⇒ y = 30
Speed of car starting from A = 50 km/hr
and from B = 30 km/hr

Question 35.
A sailor goes 8 km downstream in 40 minutes and returns in 1 hour. Find the speed of the sailor in still water and the speed of the current.
Solution:
Distance = 8 km
Let speed of sailor in still water = x km/hr
and speed of water = y km/hr
According to the conditions,

Question 36.
A boat goes 12 km upstream and 40 km downstream in 8 hours. It can go 16 km upstream and 32 km downstream in the same time. Find the speed of the boat in still water and the speed of the stream.
Solution:
Let speed of a boat = x km/hr
and speed of stream = y km/hr
According to the conditions,

Question 37.
2 men and 5 boys can finish a piece of work in 4 days, while 3 men and 6 boys can finish it in 3 days. Find the time taken by one man alone to finish the work and that taken by one boy alone to finish the work.
Solution:
Let a man can do a work in x days
His 1 day’s work =
and a boy can do a work in y days
His 1 day’s work =
According to the conditions,

Question 38.
The length of a room exceeds its breadth by 3 metres. If the length is increased by 3 metres and the breadth is decreased by 2 metres, the area remains the same. Find the length and the breadth of the room.
Solution:
Let length of a room = x m
and breadth = y m
and area = xy m²
According to the conditions,
x = y + 3 …(i)
(x + 3) (y – 2) = xy
xy – 2x + 3y – 6 = xy
-2x + 3y = 6 …(ii)
-2 (y + 3) + 3y = 6 [From (i)]
-2y – 6 + 3y = 6
⇒ y = 6 + 6 = 12
x = y + 3 = 12 + 3 = 15 …(ii)
Length of room = 15 m
and breadth = 12 m

Question 39.
The area of a rectangle gets reduced by 8 m², when its length is reduced by 5 m and its breadth is increased by 3 m. If we increase the length by 3 m and breadth by 2 m, the area is increased by 74 m². Find the length and the breadth of the rectangle.
Solution:
Let length of a rectangle = x m
and breadth = y m
Then area = x x y = xy m²
According to the conditions,
(x – 5) (y + 3) = xy – 8
⇒ xy + 3x – 5y – 15 = xy – 8
⇒ 3x – 5y = -8 + 15 = 7 …..(i)
and (x + 3) (y + 2) = xy + 74
⇒ xy + 2x + 3y + 6 = xy + 74
⇒ 2x + 3y = 74 – 6 = 68 …(ii)
Multiply (i) by 3 and (ii) by 5

Question 40.
The area of a rectangle gets reduced by 67 square metres, when its length is increased by 3 m and breadth is decreased by 4 in. If the length is reduced by 1 m and breadth is increased by 4 m, the area is increased by 89 square metres. Find the dimensions of the rectangle.
Solution:
Let length of a rectangle = x m
and breadth = y m
Then area = xy m²
According to the conditions,
(x + 3) (y – 4) = xy – 67
⇒ xy – 4x + 3y – 12 = xy – 67
⇒ -4x + 3y = -67 + 12 = -55
⇒ 4x – 3y = 55 …(i)
and (x – 1) (y + 4) = xy + 89
⇒ xy + 4x – y – 4 = xy + 89
⇒ 4x – y = 89 + 4 = 93 ….(ii)
⇒ y = 4x – 93
Substituting the value of y in (i),
4x – 3(4x – 93) = 55
⇒ 4x – 12x + 279 = 55
⇒ -8x = 55 – 279 = -224
⇒ x = 28
and y = 4x – 93 = 4 x 28 – 93 = 112 – 93 = 19
Length of rectangle = 28 m
and breadth = 19 m

Question 41.
A railway half ticket costs half the full fare and the reservation charge is the same on half ticket as on full ticket. One reserved first class ticket from Mumbai to Delhi costs ₹ 4150 while one full and one half reserved first class tickets cost ₹ 6255. What is the basic first class full fare and what is the reservation charge? [HOTS]
Solution:
Let reservation charges = ₹ x
and cost of full ticket from Mumbai to Delhi
According to the conditions,
x + y = 4150 …(i)
2x + y = 6255
⇒ 4x + 3y = 12510 …(ii)
From (i), x = 4150 – y
Substituting the value of x in (ii),
4 (4150 – y) + 3y = 12510
⇒ 16600 – 4y + 3y = 12510
-y = 12510 – 16600
-y = -4090
⇒ y = 4090
and x = 4150 – y = 4150 – 4090 = 60
Reservation charges = ₹ 60
and cost of 1 ticket = ₹ 4090

Question 42.
Five years hence, a man’s age will be three times the age of his son. Five years ago, the man was seven times as old as his son. Find their present ages.
Solution:
Let present age of a man = x years
and age of a son = y years
5 year’s hence,
Man’s age = x + 5 years
and son’s age = y + 5 years
x + 5 = 3 (y + 5) = 3y + 15
⇒ x – 3y = 15 – 5 = 10
x = 10 + 3y …(i)
and 5 years ago,
Man’s age = x – 5 years
and son’s age = y – 5 years
x – 5 = 7 (y – 5) = 7y – 35
x = 7y – 35 + 5 = 7y – 30 …(ii)
From (i) and (ii),
10 + 3y = 7y – 30
⇒ 7y – 3y = 10 + 30
⇒ 4y = 40
⇒ y = 10
and x = 10 + 3y = 10 + 3 x 10 = 10 + 30 = 40
Present age of a man = 40 years
and of son’s age = 10 years

Question 43.
Two years ago, a man was five times as old as his son. Two years later, his age will be 8 more than three times the age of his son. Find their present ages. [CBSE 2008]
Solution:
Let present age of a man = x years
and age of his son = y years
2 years ago,
Man’s age = x – 2 years
Son’s age = y – 2 years
x – 2 = 5 (y – 2)
⇒ x – 2 = 5y – 10
x = 5y – 10 + 2 = 5y – 8 …(i)
2 years later,
Man’s age = x + 2 years
and son’s age = y + 2 years
x + 2 = 3(y + 2) + 8
x + 2 = 3y + 6 + 8
⇒ x = 3y + 6 + 8 – 2 = 3y + 12 …(ii)
From (i) and (ii),
5y – 8 = 3y + 12
⇒ 5y – 3y = 12 + 8
⇒ 2y = 20
⇒ y = 10
and x = 5y – 8 = 5 x 10 – 8 = 50 – 8 = 42
Present age of man = 42 years
and age of son = 10 years

Question 44.
If twice the son’s age in years is added to the father’s age, the sum is 70. But, if twice the father’s age is added to the son’s age, the sum is 95. Find the ages of father and son.
Solution:
Let age of father = x years
and age of his son = y years
According to the conditions,
2y + x = 10 …(i)
2x + y = 95 …(ii)
From (i),
x = 70 – 2y
Substituting the value of x in (ii),
2 (70 – 2y) + y = 95
⇒ 140 – 4y + y = 95
⇒ -3y = 95 – 140 = -45
⇒ -3y = -45
⇒ y = 15
and x = 70 – 2y = 70 – 2 x 15 = 70 – 30 = 40
Age of father = 40 years
and age of his son = 15 years

Question 45.
The present age of a woman is 3 years more than three times the age of her daughter. Three years hence, the woman’s age will be 10 years more than twice the age of her daughter. Find their present ages.
Solution:
Let present age of a woman = x years
and age of her daughter = y years
According to the conditions,
x = 3y + 3 …(i)
3 years hence,
Age of woman = x + 3 years
and age of her daughter = y + 3 years
x + 3 = 2 (y + 3) + 10
⇒ x + 3 = 2y + 6 + 10
⇒x = 2y + 16 – 3 = 2y + 13 …(ii)
From (i),
3y + 3 = 2y + 13
⇒ 3y – 2y = 13 – 3
⇒ y = 10
and x = 3y + 3 = 3 x 10 + 3 = 30 + 3 = 33
Present age of woman = 33 years
and age of her daughter = 10 years

Question 46.
On selling a tea set at 5% loss and a lemon set at 15% gain, a crockery seller gains ₹ 7. If he sells the tea set at 5% gain and the lemon set at 10% gain, he gains ₹ 13. Find the actual price of each of the tea set and the lemon set. [HOTS]
Solution:
Let cost price of tea set = ₹ x
and of lemon set = ₹ y
According to the conditions,

Question 47.
A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Mona paid ₹ 21 for a book kept for 7 days, while Tanvy paid ₹ 21 for the book she kept for 5 days. Find the fixed charge and the charge for each extra day.
Solution:
Let fixed charges = ₹ x (for first three days)
and then additional charges for each day = ₹ y
According to the conditions,
Mona paid ₹ 27 for 7 dyas
x + (7 – 3) x y = 27
⇒ x + 4y = 27
and Tanvy paid ₹ 21 for 5 days
x + (5 – 3) y = 21
⇒ x + 2y = 21 …(ii)
Subtracting,
2y = 6 ⇒ y = 3
But x + 2y = 21
⇒ x + 2 x 3 = 21
⇒ x + 6 = 21
⇒ x = 21 – 6 = 15
Fixed charges = ₹ 15
and additional charges per day = ₹ 3

Question 48.
A chemist has one solution containing 50% acid and a second one containing 25% acid. How much of each should be used to make 10 litres of a 40% acid solution? [HOTS]
Solution:
Let x litres of 50% solution be mixed with y litres of 25% solution, then
x + y = 10 …(i)

Subtracting (i) from (ii),
x = 6
and x + y = 10
⇒ 6 + y = 10
⇒ y = 10 – 6 = 4
50% solution = 6 litres
and 25% solution = 4 litres

Question 49.
A jeweller has bars of 18-carat gold and 12- carat gold. How much of each must be melted together to obtain a bar of 16-carat gold, weighing 120 g? (Given : Pure gold is 24-carat). [HOTS]
Solution:
Let x g of 18 carat be mixed with y g of 12 carat gold to get 120 g of 16 carat gold, then
x + y = 120 …(i)
Now, gold % in 18-carat gold = x 100 = 75%

⇒ 3x + 2y = 320 …(ii)
From (i),
x = 120 – y
Substituting the value of x in (ii),
3 (120 – y) + 2y = 320
⇒ 360 – 3y + 2y = 320
⇒ -y = 320 – 360
⇒ -y = -40
⇒ y = 40
and 40 + x = 120
⇒ x = 120 – 40 = 80
Hence, 18 carat gold = 80 g
and 12-carat gold = 40 g

Question 50.
90% and 97% pure acid solutions are mixed to obtain 21 litres of 95% pure acid solution. Find the quantity of each type of acids to be mixed to form the mixture. [HOTS]
Solution:
Let x litres of 90% pure solution be mixed with y litres of 97% pure solution to get 21 litres of 95% pure solution. Then,
x + y = 21 …(i)

⇒ 90x + 97y = 1995
From (i), x = 21 – y
Substituting the value of x in (ii),
90 (21 – y) + 97y = 1995
⇒ 1890 – 90y + 97y = 1995
⇒ 7y = 1995 – 1890 = 105
⇒ y =15
and x = 21 – y = 21 – 15 = 6
90% pure solution = 6 litres
and 97% pure solution = 15 litres

Question 51.
The larger of the two supplementary angles exceeds the smaller by 18°. Find them.
Solution:
Let larger supplementary angle = x°
and smaller angle = y°
According to the conditions,
x + y = 180° …(i)
x = y + 18° …(ii)
From (i),
y + 18° + y = 180°
⇒ 2y = 180° – 18° = 162°
⇒ 2y = 162°
⇒ y = 81°
and x= 180°- 81° = 99°
Hence, angles are 99° and 81°

Question 52.
In a ∆ABC, ∠A = x°, ∠B = (3x – 2)°, ∠C = y° and ∠C – ∠B = 9°. Find the three angles.
Solution:
In ∆ABC,
∠A = x, ∠B = (3x – 2)°, ∠C = y°, ∠C – ∠B = 9°

Question 53.
In a cyclic quadrilateral ABCD, it is given that ∠A = (2x + 4)°, ∠B = (y + 3)°, ∠C = (2y + 10)° and ∠D = (4x – 5)°. Find the four angles.
Solution:
In a cyclic quadrilateral ABCD,

EXERCISE 3F

Very-Short and Short-Answer Questions
Question 1.
Write the number of solutions of the following pair of linear equations : x + 2y – 8 = 0, 2x + 4y = 16. [CBSE 2009]
Solution:

Question 2.
Find the value of k for which the following pair of linear equations have infinitely many solutions:
2x + 3y = 7, (k – 1) x + (k + 2) y = 3k. [CBSE 2010]
Solution:

Question 3.
For what value of k does the following pair of linear equations have infinitely many solutions?
10x + 5y – (k – 5) = 0 and 20x + 10y – k = 0.
Solution:

Question 4.
For what value of k will the following pair of linear equations have no solution?
2x + 3y = 9, 6x + (k – 2) y = (3k – 2). [CBSE 2010]
Solution:

Question 5.
Write the number of solutions of the following pair of linear equations:
x + 3y – 4 = 0 and 2x + 6y – 7 = 0.
Solution:

Question 6.
Write the value of k for which the system of equations 3x + ky = 0, 2x – y = 0 has a unique solution.
Solution:

Question 7.
The difference between two numbers is 5 and the difference between their squares is 65. Find the numbers.
Solution:
Let first, number = x
and second number = y
x – y = 5

Question 8.
The cost of 5 pens and 8 pencils is f 120, while the cost of 8 pens and 5 pencils is ₹ 153. Find the cost of 1 pen and that of 1 pencil.
Solution:
Let cost of one pen = ₹ x
and cost of one pencil = ₹ y
According to the conditions,
5x + 8y = 120 …(i)
8x + 5y = 153 …(ii)
Adding, we get
13x + 13y = 273
x + y = 21 …(iii) (Dividing by 13)
and subtracting (i) from (ii),
3x – 3y = 33
⇒ x – y = 11 …….(iv) (Dividing by 3)
Again adding (iii) and (iv),
2x = 32 ⇒ x = 16
Subtracting,
2y = 10 ⇒ y = 5
Cost of 1 pen = ₹ 16
and cost of 1 pencil = ₹ 5

Question 9.
The sum of two numbers is 80. The larger number exceeds four times the smaller one by 5. Find the numbers.
Solution:
Let first number = x
and second number = y
According to the conditions,

and x + y = 80
⇒ x + 15 = 80
x = 80 – 15 = 65
Numbers are : 65, 15

Question 10.
A number consists of two digits whose sum is 10. If 18 is subtracted from the number, its digits are reversed. Find the number.
Solution:
Let one’s digit of a two digits number = x
and ten’s digit = y
Number = x + 10y
By reversing its digits One’s digit = y
and ten’s digit = x
Then number = y + 10x
According to the conditions,
x + y = 10 …(i)
x + 10y – 18 = y + 10x
x+ 10y – y – 10x = 18
⇒ -9x + 9y = 18
⇒ x – y = -2 (Dividing by -9) …..(ii)
Adding (i) and (ii),
2x = 8 ⇒ x = 4
and by subtracting,
2y = 12 ⇒ y = 6
Number = x + 10y = 4 + 10 x 6 = 4 + 60 = 64

Question 11.
A man purchased 47 stamps of 20 p and 25 p for ₹10. Find the number of each type of stamps.
Solution:
Let number of stamps of 20p = x
and stamps of 25 p = y
According to the conditions,
x + y = 47 …..(i)
20x + 25y = 1000
4x + 5y = 200 …(ii)
From (i), x = 47 – y
Substituting the value of x in (ii),
4 (47 – y) + 5y = 200
188 – 4y + 5y = 200
⇒ y = 200 – 188 = 12
and x + y = 47
⇒ x + 12 = 47
⇒ x = 47 – 12 = 35
Hence, number of stamps of 20 p = 35
and number of stamps of 25 p = 12

Question 12.
A man has some hens and cows. If the number of heads be 48 and number of feet be 140, how many cows are there?
Solution:
Let number of hens = x
and number of cows = y
According to the conditions,
x + y = 48 …..(i)
x x 2 + y x 4 = 140
⇒ 2x + 4y = 140
⇒ x + 2y = 70 ……(ii)
Subtracting (i) from (ii),
y = 22
and x + y = 48
⇒ x + 22 = 48
⇒ x = 48 – 22 = 26
Number of hens = 26
and number of cows = 22

Question 13.

Solution:

Question 14.

Solution:

Question 15.
If 12x + 17y = 53 and 17x + 12y = 63 then find the value of (x + y).
Solution:
12x + 17y = 53 …(i)
17x + 12y = 63 …(ii)
Adding, 29x + 29y = 116
Dividing by 29,
x + y = 4 …(iii)
Subtracting,
-5x + 5y = -10
⇒ x – y = 2 …(iv) (Dividing by -5)
Adding (iii) and (iv)
2x = 6 ⇒ x = 3
Subtracting,
2y = 2 ⇒ y = 1
x = 3, y = 1
x + y = 3 + 1 = 4

Question 16.
Find the value of k for which the system 3x + 5y = 0, kx + 10y = 0 has a nonzero solution.
Solution:

Question 17.
Find k for which the system kx – y = 2 and 6x – 2y = 3 has a unique solution.
Solution:
kx – y = 2
6x – 2y = 3

Question 18.
Find k for which the system 2x + 3y – 5 = 0, 4x + ky – 10 = 0 has an infinite number of solutions.
Solution:

Question 19.
Show that the system 2x + 3y – 1 = 0, 4x + 6y – 4 = 0 has no solution.
Solution:

Question 20.
Find k for which the system x + 2y = 3 and 5x + ky + 7 = 0 is inconsistent.
Solution:

Question 21.
Solve:

Solution:

MULTIPLE CHOICE QUESTIONS

Choose the correct answer in each of the following questions.
Question 1.
If 2x + 3y = 12 and 3x – 2y = 5 then
(a) x = 2, y = 3
(b) x = 2, y = -3
(c) x = 3, y = 2
(d) x = 3, y = -2
Solution:
(c)

Question 2.
If x – y = 2 and then
(a) x = 4, y = 2
(b) x = 5, y = 3
(c) x = 6, y = 4
(d) x = 7, y = 5
Solution:
(c)

Question 3.

Solution:
(a)

Question 4.

Solution:
(d)

Question 5.

Solution:
(a)

Question 6.

Solution:
(b)

Question 7.
If 4x + 6y = 3xy and 8x + 9y = 5xy then
(a) x = 2, y = 3
(b) x = 1, y = 2
(c) x = 3, y = 4
(d) x = 1, y = -1
Solution:
(c) 4x + 6y = 3xy, 8x + 9y = 5xy
Dividing each term by xy,

Question 8.
If 29x + 37y = 103 and 37x + 29y = 95 then
(a) x = 1, y = 2
(b) x = 2, y = 1
(c) x = 3, y = 2
(d) x = 2, y = 3
Solution:
(a)

Question 9.

(c) 0
(d) none of these
Solution:
(c)

Question 10.

Solution:
(b)

Question 11.
The system kx – y = 2 and 6x – 2y = 3 has a unique solution only when
(a) k = 0
(b) k ≠ 0
(c) k = 3
(d) k ≠ 3
Solution:
(d)

Question 12.
The system x – 2y = 3 and 3x + ky = 1 has a unique solution only when
(a) k = -6
(b) k ≠ -6
(c) k = 0
(d) k ≠ 0
Solution:
(b)

Question 13.
The system x + 2y = 3 and 5x + ky + 7 = 0 has no solution, when
(a) k =10
(b) k ≠ 10
(c) k =
(d) k = -21
Solution:
(a)

Question 14.
If the lines given by 3x + 2ky = 2 and 2x + 5y + 1 = 0 are parallel then the value of k is
(a)
(b)
(c)
(d)
Solution:
(d)

Question 15.
For what value of k do the equations kx – 2y = 3 and 3x + y = 5 represent two lines intersecting at a unique point?
(a) k = 3
(b) k = -3
(c) k = 6
(d) all real values except -6
Solution:
(d)

Question 16.
The pair of equations x + 2y + 5 = 0 and -3x – 6y + 1 = 0 has
(a) a unique solution
(b) exactly two solutions
(c) infinitely many solutions
(d) no solution
Solution:
(d)

Question 17.
The pair of equations 2x + 3y = 5 and 4x + 6y = 15 has
(a) a unique solution
(b) exactly two solutions
(c) infinitely many solutions
(d) no solution
Solution:
(d)

Question 18.
If a pair of linear equations is consistent then their graph lines will be
(a) parallel
(b) always coincident
(c) always intersecting
(d) intersecting or coincident
Solution:
(d) The system of equations is consistent then their graph lines will be either intersecting or coincident.

Question 19.
If a pair of linear equations is inconsistent then their graph lines will be
(a) parallel
(b) always coincident
(c) always intersecting
(d) intersecting or coincident
Solution:
(a) The pair of lines of equation is inconsistent then the system will not have no solution i.e., their lines will be parallel.

Question 20.
In a ∆ABC, ∠C = 3∠B = 2 (∠A + ∠B), then ∠B = ?
(a) 20°
(b) 40°
(c) 60°
(d) 80°
Solution:
(b)

Question 21.
In a cyclic quadrilateral ABCD, it is being given that ∠A = (x + y + 10)°, ∠B = (y + 20)°, ∠C = (x + y – 30)° and ∠D = (x + y)°. Then, ∠B = ?
(a) 70°
(b) 80°
(c) 100°
(d) 110°
Solution:
(b) ABCD is a cyclic quadrilateral
∠A = (x + y + 10)°, ∠B = (y + 20)°, ∠C = (x + y – 30)° and ∠D = (x + y)°
∠A + ∠C = 180°
Now, x + y + 10°+ x + y – 30° = 180°
⇒ 2x + 2y – 20 = 180°
⇒ 2x + 2y = 180° + 20° = 200°
⇒ x + y = 100° …(i)
and ∠B + ∠D = 180°
⇒ y + 20° + x + y = 180°
⇒ x + 2y = 180° – 20° = 160° …(ii)
Subtracting,
-y = -60° ⇒ y = 60°
and x + 60° = 100°
⇒ x = 100° – 60° = 40°
Now, ∠B = y + 20° = 60° + 20° = 80°

Question 22.
The sum of the digits of a two-digit number is 15. The number obtained by interchanging the digits exceeds the given number by 9. The number is
(a) 96
(b) 69
(c) 87
(d) 78
Solution:
(d) Let one’s digit of a two digit number = x
and ten’s digit = y
Number = x + 10y
By interchanging the digits,
One’s digit = y
and ten’s digit = x
Number = y + 10x
According to the conditions,
x + y = 15 …(i)
y + 10x = x + 10y + 9
⇒ y + 10x – x – 10y = 9
⇒ 9x – 9y = 9
⇒ x – y = 1 …(ii)
Adding (i) and (ii),
2x = 16 ⇒ x = 8
and x + y = 15
⇒ 8 + y = 15
⇒ y = 15 – 8 = 7
Number = x + 10y = 8 + 10 x 7 = 8 + 70 = 78

Question 23.
In the given fraction, if 1 is subtracted from the numerator and 2 is added to the denominator, it becomes . If 7 is subtracted from the numerator and 2 is subtracted from the denominator, it becomes . The fraction is
(a)
(b)
(c)
(d)
Solution:
(b) Let the numerator of a fractions = x
and denominator = y

Question 24.
5 years hence, the age of a man shall be 3 times the age of his son while 5 years earlier the age of the man was 7 times the age of his son. The present age of the man is
(a) 45 years
(b) 50 years
(c) 47 years
(d) 40 years
Solution:
(d) Let present age of man = x years
and age of his son = y years
5 years hence,
Age of man = (x + 5) years
and age of son = y + 5 years
(x + 5) = 3 (y + 5)
⇒ x + 5 = 3y + 15
x = 3y + 15 – 5
x = 3y + 10 ……(i)
and 5 years earlier
Age of man = x – 5 years
and age of son = y – 5 years
x – 5 = 7 (y – 5)
x – 5 = 7y – 35
⇒ x = 7y – 35 + 5
x = 7y – 30 ……….(ii)
From (i) and (ii),
7y – 30 = 3y + 10
⇒ 7y – 3y = 10 + 30
⇒ 4y = 40
y = 10
x = 3y + 10 = 3 x 10 + 10 = 30 + 10 = 40
Present age of father = 40 years

Question 25.
The graphs of the equations 6x – 27 + 9 = 0 and 3x – 7 + 12 = 0 are two lines which are
(a) coincident
(b) parallel
(c) intersecting exactly at one point
(d) perpendicular to each other
Solution:
(b)

Question 26.
The graphs of the equations 2x + 3y – 2 = 0 and x – 27 – 8 = 0 are two lines which are
(a) coincident
(b) parallel
(c) intersecting exactly at one point
(d) perpendicular to each other
Solution:
(c)

Question 27.
The graphs of the equations 5x – 15y = 8 and 3x – 9y = are two lines which are
(a) coincident
(b) parallel
(c) intersecting exactly at one point
(d) perpendicular to each other
Solution:
(a)

The system has infinitely many solutions.
The lines are coincident.

TEST YOURSELF

Question 1.
The graphic representation of the equations x + 2y = 3 and 2x + 4y + 1 = 0 gives a pair of
(a) parallel lines
(b) intersecting lines
(c) coincident lines
(d) none of these
Solution:
(a)

Question 2.
If 2x – 3y = 7 and (a + b) x – (a + b -3) y = 4a + b have an infinite number of solutions then
(a) a = 5, b = 1
(b) a = -5, b = 1
(c) a = 5, b = -1
(d) a = -5, b = -1
Solution:
(d)

Question 3.
The pair of equations 2x + y = 5, 3x + 2y = 8 has
(a) a unique solution
(b) two solutions
(c) no solutions
(d) infinitely many solutions
Solution:
(a)

Question 4.
If x = -y and y > 0, which of the following is wrong?
(a) x²y > 0
(b) x + y = 0
(c) xy < 0
(d) – = 0
Solution:
(d)

Short-Answer Questions
Question 5.
Show that the system of equations -x + 2y + 2 = 0 and x – y – 1 = 0 has a unique solution.
Solution:

Question 6.
For what values of k is the system of equations kx + 3y = k – 2, 12x + ky = k inconsistent?
Solution:

Question 7.

Solution:

Question 8.
Solve the system of equations x – 2y = 0, 3x + 4y = 20.
Solution:

Question 9.
Show that the paths represented by the equations x – 3y = 2 and -2x + 6y = 5 are parallel.
Solution:

Question 10.
The difference between two numbers is 26 and one number is three times the other. Find the numbers.
Solution:
Let first number = x
and second number = y
According to the conditions, x – y = 26 …(i)
and x = 3y …..(ii)
From (i),
3y – y = 26
⇒ 2y = 26
⇒ y = 13
and x = 3 x 13 = 39
Numbers are 39 and 13

Short-Answer Questions (3 marks)
Question 11.
Solve : 23x + 29y = 98, 29x + 23y = 110.
Solution:
23x + 29y = 98 …..(i)
29x + 23y = 110 …..(ii)
Adding, we get 52x + 52y = 208
x + y = 4 …..(iii) (Dividing by 52)
and subtracting,
-6x + 6y = -12
x – y = 2. …..(iv) (Dividing by -6)
Adding (iii) and (iv),
2x = 6 ⇒ x = 3
Subtracting,
2x = 2 ⇒ y = 1
Hence, x = 3, y = 1

Question 12.
Solve : 6x + 3y = 7xy and 3x + 9y = 11xy.
Solution:


x = 1, y =

Question 13.
Find the value of k for which the system of equations 3x + y = 1 and kx + 2y = 5 has (i) a unique solution, (ii) no solution.
Solution:

Question 14.
In a ∆ABC, ∠C = 3∠B = 2 (∠A + ∠B). Find the measure of each one of ∠A, ∠B and ∠C.
Solution:

Question 15.
5 pencils and 7 pens together cost ₹ 195 while 7 pencils and 5 pens together cost ₹ 153. Find the cost of each one of the pencil and the pen.
Solution:
Let cost of one pencil = ₹ x
and cost of one pen = ₹ y
According to the condition,
5x + 7y = 195 …(i)
7x + 5y= 153 …(ii)
Adding, (i) and (ii)
12x + 12y = 348
x + y = 29 ….(iii) (Dividing by 12)
and subtracting,
-2x + 2y = 42
-x + y = 21 …..(iv) (Dividing by -2)
Now, Adding (iii) and (iv),
2y = 50 ⇒ y = 25
and from (iv),
-x + 25 = 21 ⇒ -x = 21 – 25 = -4
x = 4
Cost of one pencil = ₹ 4
and cost of one pen = ₹ 25

Question 16.
Solve the following system of equations graphically:
2x – 3y = 1, 4x – 3y + 1 = 0.
Solution:
2x – 3y = 1, 4x – 3y + 1 = 0
2x – 3y = 1
2x = 1 + 3y
x =
Giving some different values to y, we get corresponding values of x as given below

Now plot the points (2, 1), (5, 3) and (-1, -1) on the graph and join them to get a line.
Similarly,
4x – 3y + 1 = 0
⇒ 4x = 3y – 1
⇒ x =

Now plot the points (-1, -1), (-4, -5) and (2, 3) on the graph and join them to get another line which intersects the first line at the point (-1, -1).
Hence, x = -1, y = -1

Long-Answer Questions
Question 17.
Find the angles of a cyclic quadrilateral ABCD in which ∠A = (4x + 20)°, ∠B = (3x – 5)°, ∠C = (4y)° and ∠D = (7y + 5)°.
Solution:
We know that opposite angles of a cyclic quadrilateral are supplementary.
∠A + ∠C = 180° and ∠B + ∠D = 180°
Now, ∠A = 4x° + 20°, ∠B = 3x° – 5°, ∠C = 4y° and ∠D = 7y° + 5°
But ∠A + ∠C = 180°
4x + 20° + 4y° = 180°
⇒ 4x + 4y = 180° – 20 = 160°
x + y = 40° …(i) (Dividing by 4)
and ∠B + ∠D = 180°
⇒ 3x – 5 + 7y + 5 = 180°
⇒ 3x + 7y = 180° …(ii)
From (i), x = 40° – y
Substituting the value of x in (ii),
3(40° – y) + 7y = 180°
⇒ 120° – 3y + 7y = 180°
⇒ 4y = 180°- 120° = 60°
y = 15°
and x = 40° – y = 40° – 15° = 25°
∠A = 4x + 20 = 4 x 25 + 20 = 100 + 20= 120°
∠B = 3x – 5 = 3 x 25 – 5 = 75 – 5 = 70°
∠C = 4y = 4 x 15 = 60°
∠D = 7y + 5 = 7 x 15 + 5 = 105 + 5 = 110°

Question 18.
Solve for x and y :

Solution:

Question 19.
If 1 is added to both the numerator and the denominator of a fraction, it becomes . If, however, 5 is subtracted from both numerator and the denominator, the fraction becomes . Find the fraction.
Solution:
Let numerator of a fraction = x
and denominator = y
Fraction =
According to the conditions,

Question 20.
Solve : – = a + b, ax – by = 2ab.
Solution: