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# Short Answer Type Questions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev

## Class 10 : Short Answer Type Questions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev

The document Short Answer Type Questions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev is a part of the Class 10 Course Class 10 Mathematics by VP Classes.
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Ques 1: For what value of p, the pair of linear equations px = 2y; 2x - y + 5 = 0 has unique solution?
Sol: We have:
px = 2y
⇒ p - 2y = 0
2x = y + 5
⇒ 2x - y = - 5
Here, a1 = p, b1 = - 2,
c1 = 0
a= 2, b2 = - 1,
c2 = - 5
For a unique solution,

⇒ p ≠ 2 × 2
⇒ p ≠ 4

Ques 2: In a cyclic quadrilateral PQRS, ∠P = (2x + 4)°, ∠Q = (y + 3)°, ∠R = (2y + 10)° and ∠S = (4x - 5)°. Find its four angles.
Sol:
In a cyclic quadrilateral, the opposite angles are supplementary.
∴∠P + ∠R = 180°
⇒ (2x + 4)° + (2y + 10)° = 180°
⇒ 2x + 2y + 14 - 180 = 0
⇒ 2x + 2y - 166 = 0
⇒ x + y - 83 = 0  ...(1)
Also ∠Q + ∠S = 180°
∴ (y + 3)° + (4x - 5)° = 180°
⇒ y + 4x - 2 - 180 = 0
⇒ y + 4x - 182 = 0 ...(2)
From (1) and (2),
a1 = 1, b1 = 1, c1 = - 83
a2 = 4, b2 = 1, c2 = - 182

∴ ∠P = (2x + 4)° = [(2 × 33) + 4]° = 70°
∠Q = (y + 3)° = [50 + 3]° = 53°
∠R = (2y + 10)° = [2 × 50 + 10]° = 110°
∠S = (4x - 5)° = [4 × 33 - 5°] = 127°

Ques 3: Solve:
23x + 35y = 209
35x + 23y = 197
Sol: We have:
23x + 35y = 209   ...(1)
35x + 23y = 197    ...(2)
88x + 88y = 406 [Adding (1) and (2)]
⇒ x + y = 7 ...(3) [Dividing by 88]
Subtracting (1) from (2),
35x + 23y = 197
23x + 35y = 209
(-) (-) (-)
12x - 12y = - 12
⇒ x - y = 1 ...(4) [Dividing by 12]
2x = 8 ⇒ x = 4
From (3) x + y = 7 ⇒ 4 + y = 7
⇒ y = 3
So, x = 4 and y = 3.

Ques 4: Solve:
3x + 5y = 70   ...(1)
7x - 3y = 60   ...(2)
Sol: From (1) and (2), we have:
a1 = 3, b1 = 5, c1 = - 70
a2 = 7, b2 = - 3, c2 = - 60

Thus, the required solution is:
.

Ques 5: Without drawing the graphs, state whether the following pair of linear equations will represent intersecting lines, coinciding lines or parallel lines:
6x - 3y + 10 = 0
2x - y + 9 = 0
Sol: Here, the given set of equations is:
6x - 3y + 10 = 0   ...(1)
2x - y + 9 = 0    ...(2)
From (1) and (2), we have:
a1 = 6, b1 = - 3, c1 = 10
a2 = 2, b2 = - 1, c2 = 9
Now,
∴We have
=
This condition represents parallel lines. Hence, the given pair represents parallel lines.

Ques 6: Check graphically whether the pair of equations
3x - 2y + 2 = 0
= 0
is consistent. Also find the co-ordinates of the points where the graphs of the equations meet the y-axis.
Sol: ∴ 3x - 2y + 2 = 0 ⇒ y =  ..(1)

Also = 0 ⇒ y =  ...(2)

Plotting the points (0, 1), (2, 4), (- 2, - 2) and (0, 3), (2, 6), (–2, 0) we get two straight lines l1 and l2 which are parallel.
∴ The given equations are inconsistent.
From the graph, we observe that line l1 meets the y-axis at (0, 1) and line l2 meets the y-axis at (0, 3).

Ques 7: A fraction becomes 1/3 if 2 is added to both numerator and denominator. If 3 is added to both numerator and denominator, it becomes 2/5. Find the fraction.
Sol: Let the fraction be x/y.
From 1st condition,

⇒ 3x + 6 = y + 2
⇒ 3x − y + 4 = 0 ...(1)
From 2nd conditon,

⇒ 5x + 15 = 2y + 6
⇒ 5x − 2y + 9 = 0 ...(2)
From (1) and (2), we have:

Ques 8: Check graphically whether the pair of equations
3x + 5y = 15
x - y = 5
is consistent. Also, find the coordinates of the points where the graphs of equations meet the y-axis.
Sol: We have
3x + 5y = 15

And from x - y = 5
⇒ y = x - 5

Plotting the above two sets of points we get two straight lines l1 and l2 which intersect at the point (5, 0).
Thus, the given system is consistent.

Obviously, line l1 meets the y-axis at (0, 3) and line l2 meets the y-axis at (0, - 5).

Ques 9: Places A and B are 160 km apart on the highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 8 hours, but if they travel towards each other, they meet in 2 hours. What are the speeds of the two cars?
Sol: Let the car-I and car-II start from A and B at x km/hr and y km/hr respectively.
Case-I: [Cars are moving in the same direction]

Let the two cars meet at C after 8 hours.
Distance covered:
by car-I = AC = 8x km
by car-II = BC = 8y km
∴ AB = AC - BC
⇒ 160 = 8x - 8y
⇒ x - y = 20   ...(1)
Case-II: [Cars are moving in opposite directions]

Let, after 2 hours, the cars meet at D.
∴ Distance cover after 2 hours,
by car-I = AD = 2x km
by car-II = BD = 2y km
⇒ AB = AD + BD
⇒ 160 = 2x + 2y
⇒ 80 = x + y
⇒ x + y = 80 ...(2)
Adding (1) and (2), we get
x + y = 80
x - y = 20
2x = 100
⇒ x = 100/2 = 50
⇒ Substituting x = 50 in (1), we get
x - y = 20 ⇒ 50 - y = 20
⇒ y = 50 - 20 = 30
⇒ Speed of car-I = 50 km/hr
Speed of car-II = 30 km/hr.

Ques 10: Solve for x and y:

Sol: We have:
=    ...(1)
ax − by = 2ab    ...(2)
Dividing (2) by a, we have:

...(3)
From (1) and (3), we have

From (2),
ab - by = 2ab
⇒ - by = 2ab - ab = ab

y = - a
Thus, x = b and y = - a.

Ques 11: The sum of two numbers is 8. Determine the numbers if the sum of their reciprocals is 8/15.
Sol: Let the two numbers be x and y.
According to the conditions:
x + y = 8 ...(1)
= ..(2)
From (1), x = (8 - y)
Substituting x = (8 - y) in (2),

⇒ 8 × 15 = 8 × y (8 - y)
⇒ 64y - 8y2 - 120 = 0
⇒- y2 + 8y - 15 = 0
⇒ y2 - 8y + 15 = 0
⇒ y2 - 5y - 3y + 15 = 0
⇒ y (y - 5) - 3 (y - 5) = 0
⇒ (y - 5) (y - 3) = 0
⇒ If y - 5 = 0 then y = 5
or if y - 3 = 0, then y = 3
Since x = 8 - y
⇒ when y = 5,
then x = 8 - 5 = 3
when y = 3, then
x = 8 - 3 = 5
⇒ The required numbers are (3, 5) or
(5, 3).

Ques 12: Solve the following pair of equations:

Sol: Let
⇒The given system of equations becomes:
5p + q = 2 ...(1)
6p - 3q = 1 ...(2)
Multiplying (1) by 3 and adding to (2),

and 5p + q = 2 ⇒

Since,

= 3 ⇒ x = 4
Also,
⇒ y - 2 = 3 ⇒ y = 5
Thus, x = 4 and y = 4
⇒ 3y - 6 = 1
⇒ 3y = 1 + 6 = 7
⇒ y = 7/3

Ques 13: Solve the following pair of equations:

Sol: Let
⇒ The given pair of equation is expressed as
10p + 2q = 4 ⇒ 5p + q = 2   ...(1)
15p - 5q = - 2 ...(2)
Multiplying (1) by 5 and adding to (2)

From (1),
⇒1 + q = 2 ⇒ q = 2 - 1 = 1
Since,

⇒ x + y = 5   ...(3)
And
⇒ x − y = 1   ...(4)

From (3), 3 + y = 5 ⇒ y = 2
Thus, x = 3 and y = 2.

Ques 14: Solve for x and y:
37x + 43y = 123
43x + 37y = 117
Sol: We have:
37x + 43y = 123 ...(1)
43x + 37y = 117 ...(2)

Dividing both sides by 80, we get
x + y = 3 ...(3)
Subtracting (2) from (1),
- 6x + 6y = 6 ⇒ - x + y = 1 ...(4)

⇒ y = 4/2 = 2
Putting y = 2 in x + y = 3, we get
x + 2 = 3 ⇒ x = 3 - 2 = 1
Thus, x = 1 and y = 2.

Ques 15: Solve for ‘x’ and ‘y’:
(a - b) x + (a + b) y = a2 - 2ab - b2
(a + b) (x + y) = a2 + b2
Sol: We have:
(a - b) x + (a + b) y = a2 - 2ab - b2   ...(1)
(a + b) x + (a + b) y = a2 + b2   ...(2)

− 2b x = − 2ab − b
⇒ (− 2b) x = − 2b (a + b)

From (2),
(a + b) (a + b) + (a + b) y = a2 + b2
⇒(a + b)2 + (a + b) y = (a2 + b2)
⇒ (a + b) y = (a2 + b2) - (a + b)2
⇒ (a + b) y = a2 + b2 - (a2 + b2 + 2ab)
⇒ (a + b) y = a2 + b2 - a2 - b2 - 2ab
⇒ (a + b) y = - 2ab

Thus, x = (a + b) and

Ques 16: Represent the following pair of equations graphically and write the co-ordinates of points where the lines intersect y-axis:
x + 3y = 6, 2x - 3y = 12
Sol: We have:
x + 3y = 6

and 2x - 3y = 12

Plotting the above points, we get two straight lines l1 and l2 such that they intersect at (6, 0) as shown below:

Obviously,
The line l1 meets the y-axis at (0, 2).
The line l2 meets the y-axis at (0, - 4).

Ques 17: Solve for x and y:

Sol: Let
∴ We have:
5p + q = 2   ...(1)
6p - 3q = 1   ...(2)
From (1) and (2), we have:

And
Now,
⇒ x − 1 = 3 ⇒ x = 4
And
⇒ y − 2 = 3 ⇒ y = 5
Thus x = 4 and y = 5

Ques 18: For what values of ‘a’ and ‘b’ does the following pair of equations have an infinite number of solutions?
2x + 3y = 7
a (x + y) - b (x - y) = 3a + b - 2
Sol: We have:
2x + 3y = 7   ...(1)
a (x + y) - b (x - y) = 3a + b - 2   ...(2)
From (2), we have:
a (x + y) - b (x - y) = 3a + b - 2
⇒ ax + ay - bx + by = 3a + b - 2
⇒ ax - bx + ay + by = 3a + b - 2
⇒ (a - b) x + (a + b) y = 3a + b - 2
Now, A1 = 2, B1= 3,
C1 = - 7
A2 = (a - b), B2
= (a + b),
C2 = - [3a + b - 2]
For infinite number of solutions,

i.e.,

⇒ 2 (a + b) = 3 (a − b)
⇒ 2a + 2b − 3a + 3b = 0
⇒ − a + 5b = 0
⇒ a = 5b ...(3)
Also =
⇒ 3 (3a + b - 2) = 7 (a + b)
⇒ 9a + 3b - 6 = 7a + 7b
⇒ 9a - 7a + 3b - 7b = 6
⇒ 2a - 4b = 6
⇒ a - 2b = 3    ...(4)
From (3) and (4),
5b - 2b = 3
⇒ 3b = 3 ⇒ b = 1
Thus, a = 5 × b
⇒ a = 5 × 1 = 5
i.e., a = 5 and b = 1.

Ques 19: Solve the following pairs of equations for x and y:

Sol: Let
∴ We have:
15p + 22q = 5 ...(1)
40p + 55q = 13 ...(2)
From (1) and (2), we get

Adding (3) and (4), we have
2x = 16 ⇒ x = 16/2 =8
From (4), 8 + y = 11 ⇒ y = 11 - 8 = 3
Thus, x = 8 and y = 3.

Ques 20: Draw the graph of the pair of linear equations x – y + 2 = 0 and 4x – y – 4 = 0. Calculate the area of triangle formed by the lines so drawn and the x-axis.
Sol:
To draw the graph of the given pair of equations, we have the table of ordered pairs

Plot the points A(0, 2), B(–2, 0); C(0, –4) and D(1, 0) on the graph paper and join the points to form the lines AB and CD :

From the graph, we find that the points P(2, 4) is common to both the lines AB and CD.
These lines meet the x-axis at B(–2, 0) and D(1, 0).
Thus, the triangle BDP is formed by the lines and the x-axis.
The vertices of this Δ are
B(–2, 0), D(1, 0)  and    P(2, 4)
Now, the base of ΔBDP = BD
= (BO + OD)
= (2 + 1) units
= 3 units
Altitude of the ΔBDP = PQ
= 4 units
∴ Area of ΔBDP =

=  6 sq. units

Ques 21: It can take 12 hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for 4 hours and the pipe of smaller diameter for 9 hours, only half the pool can be filled. How long would it take for each pipe to fill the pool separately?
Sol: Let the time taken by the pipe of larger diameter to fill the pool separately = x hours.
The time taken by the pipe of smaller diameter to fill the pool separately = y hours.
∴ Part of the pool fill by a pipe of larger diameter in 1 hour = 1/x.
Part of fool filled by the pipe of larger diameter in 4 hours = 4/x.
Similarly, Part of the pool filled by the pipe of smaller diameter in 9 hours = 9/y.
∴We have =  ...(1)
Since the pool is filled by both the pipes together in 12 hours.
∴     ...(2)
To solve (1) and (2), multiplying (1) by 3 and subtracting (2) from it, we have

Substituting, y = 30 in (2), we have

⇒ x = 20
⇒ Required time taken by pipe of larger diameter = 20 hours

Required time taken by pipe of smaller diameter = 30 hours

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