Class 10 Maths Chapter 3 HOTS Questions - Pair of Linear Equations in Two Variables

Q1. If two of the roots of f(x) = x³ – 5x² – 16x + 80 are equal in magnitude but opposite in sign, then find all of its zeroes.

Sol: Let α and β are the two zeroes which are equal in magnitude but opposite in sign.
∴ α + β = 0
(Let the third zero is γ)
Sum of zeroes of
f(x)= α + β + γ =  -(-5)1 = 5
∴ γ = 5  [α + β = 0]
Product of zeroes

= αβγ = -801 ⇒ 5 × αβ = -80 [∵ γ = 5]

∴  αβ = -805 = -16

⇒ – α² = –16 [α + β = 0 ⇒ β = – α]

⇒ α² = 16 ⇒ α = ± 4

α = ± 4 ⇒ β = ∓ 4    

[∴ β = –α]

Thus, the zeroes are : [± 4, ∓ 4, and 5]

Q2. Solve : 

3x + 2y = 13

5x - 3y = 9

Sol : Given equations:

3x + 2y = 13

5x - 3y = 9

Let 1x = p and 1y = q

The equations become:

3p + 2q = 13 ...(i)

5p - 3q = 9 ...(ii)

From equation (i):

2q = 13 - 3p

q = 13 - 3p2

Substitute q into equation (ii):

5p - 3( 13 - 3p2 ) = 9

Multiply through by 2 to eliminate the fraction:

10p - (39 - 9p) = 18

19p = 57

p = 3

Solve for q:

q = 13 - 92 = 42 = 2

Back-substitute for x and y:

x = 1p = 13

y = 1q = 12

Final Answer:

x = 13 , y = 12

Q3. Solve :

8x + 2y + 32x − y = 3

12x + 2y − 62x − y = 1

Hint: Put x + 2y = p and 2x – y = q

Sol:  Given equations:
8x + 2y + 32x - y = 3
12x + 2y - 62x - y = 1

Let:
x + 2y = p, 2x - y = q

Substituting:
8p + 3q = 3 ...(1)
12p - 6q = 1  ...(2)

Eliminating fractions:
8q + 3p = 3pq  ...(3)
12q - 6p = pq  ...(4)

From (4):
pq = 12q - 6p

Substitute pq in (3):
8q + 3p = 3(12q - 6p)
8q + 3p = 36q - 18p
21p = 28q
p = 4q3

Substitute p into (1):
8p + 3q = 3
6q + 3q = 3
q = 3

Substitute q = 3 into p:
p = 4 × 33 = 4

Back-substitute:
x + 2y = 4, 2x - y = 3
Solving gives: x = 2, y = 1

Final Answer: x = 2, y = 1

Q4. Solve : x + y = 18 ; y + z  = 12 ;  z + x = 16.
Sol: Adding the three equations, we get

⇒ x + y + z  = 23
Now, (x + y + z = 23) – (x + y = 18)
⇒ z = 5 (x + y + z = 23) – (y + z = 12)
⇒ x = 11 (x + y + z = 23) – (z + x = 16)
⇒ y = 7  
Thus, x = 11, y = 7 and z = 5

Q5. Solve :  xyx + y = 43 , yzy + z = 125 , zxz + x = 32
Sol: Inverting the equations:

x + yxy = 34 ⇒ 1x + 1y = 34 ...(1)

y + zyz = 512 ⇒ 1y + 1z = 512 ...(2)

z + xzx = 23 ⇒ 1z + 1x = 23 ...(3)

Adding (1), (2) and (3), we get

2x + 2y + 2z = 34 + 512 + 23 = 2212 = 116 

⇒ 1x + 1y + 1z = 1112 ...(4)
Now, subtracting (1), (2) and (3) turn by turn from (4), we get x = 2, y = 4 and z = 6

Q6. Solve: x + y − 32 = x + 2y − 43 = 3x + y11
Sol: 
From x + y − 32 = 3x + y11
⇒ 11(x + y – 3) = 2(3x + y) ⇒ 5x + 9y = 33    ...(1)
From x + 2y − 43 = 3x + y11
⇒ 11(x + 2y – 4) = 3(3x + y)
⇒ 2x + 9y = 44        ...(2)
Solving (1) and (2), we get x = 3 and y = 2