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, a_{1} = p, b_{1} = - 2,

c_{1} = 0

a_{2 }= 2, b_{2} = - 1,

c_{2} = - 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),

a

a

⇒

⇒

∴ ∠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°

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]

Adding (3) and (4),

2x = 8 ⇒ x = 4

From (3) x + y = 7 ⇒ 4 + y = 7

⇒ y = 3

So, x = 4 and y = 3.

a

a

⇒

∴

Thus, the required solution is:

.

6x - 3y + 10 = 0 ...(1)

2x - y + 9 = 0 ...(2)

From (1) and (2), we have:

a

a

Now,

∴We have

=

This condition represents parallel lines. Hence, the given pair represents parallel lines.

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 l

∴ The given equations are inconsistent.

From the graph, we observe that line l

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 l_{1} meets the y-axis at (0, 3) and line l_{2} 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 - 8y^{2} - 120 = 0

⇒- y^{2} + 8y - 15 = 0

⇒ y^{2} - 8y + 15 = 0

⇒ y^{2} - 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)

Adding (3) and (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)

Adding (1) and (2)

Dividing both sides by 80, we get

x + y = 3 ...(3)

Subtracting (2) from (1),

- 6x + 6y = 6 ⇒ - x + y = 1 ...(4)

Adding:

⇒ 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 = a ^{2} - 2ab - b^{2}**

(a - b) x + (a + b) y = a

(a + b) x + (a + b) y = a

− 2b x = − 2ab − b

⇒ (− 2b) x = − 2b (a + b)

⇒

From (2),

(a + b) (a + b) + (a + b) y = a

⇒(a + b)2 + (a + b) y = (a

⇒ (a + b) y = (a

⇒ (a + b) y = a

⇒ (a + b) y = a

⇒ (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 l_{1} and l_{2} such that they intersect at (6, 0) as shown below:

Obviously,

The line l_{1} meets the y-axis at (0, 2).

The line l_{2} 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, A_{1} = 2, B_{1}= 3,

C_{1} = - 7

A_{2} = (a - b), B_{2}

= (a + b),

C_{2} = - [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

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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