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### Exercise 3.2 Pair Of Linear Equations In Two Variables Class 10th Mathematics KC Sinha Solution

Exercise 3.2
1. 3x - y = 3 9x — 3y = 9 Solve the following pair of linear equations by…
2. 7x— 15y = 2 x + 2y = 3 Solve the following pair of linear equations by…
3. x + y = 14 x - y = 4 Solve the following pair of linear equations by…
4. 0.5x + 0.8y = 3.4 0.6x — 0.3y = 0.3 Solve the following pair of linear…
5. x + y = a—b ax —by = a^2 + b^2 Solve the following pair of linear equations by…
6. x + y = 2m mx — ny = m^2 + n^2 Solve the following pair of linear equations by…
7. x/2 + y = 0.8 x + y/2 = 7/10 Solve the following system of equations by…
8. s —t = 3 s/3 + t/2 = 6 Solve the following system of equations by substitution…
9. x/a + y/b = 2 , a ≠ 0, b ≠ 0 ax - by = a^2 - b^2 Solve the following system of…
10. bx/a + ay/b = a^2 + b^2 x + y = 2ab Solve the following system of equations by…
###### Exercise 3.2
Question 1.

Solve the following pair of linear equations by substitution method:

3x – y = 3

9x — 3y = 9

Given equations are

3x – y = 3 …(i)

9x – 3y = 9 …(ii)

From eqn (i), y = 3x – 3 …(iii)

On substituting y = 3x – 3 in eqn (ii), we get

⇒ 9x – 3(3x – 3) = 9

⇒ 9x – 9x + 9 = 9

⇒ 9 = 9

This equality is true for all values of x, therefore given pair of equations have infinitely many solutions.

Question 2.

Solve the following pair of linear equations by substitution method:

7x— 15y = 2

x + 2y = 3

Given equations are

7x – 15y = 2 …(i)

x + 2y = 3 …(ii)

From eqn (ii), x = 3 – 2y …(iii)

On substituting x = 3 – 2y in eqn (i), we get

⇒ 7(3 – 2y) – 15y = 2

⇒ 21 – 14y – 15y = 2

⇒ 21 – 29y = 2

⇒ – 29y = – 19

Now, on putting  in eqn (iii), we get

Thus,  and is the required solution.

Question 3.

Solve the following pair of linear equations by substitution method:

x + y = 14

x – y = 4

Given equations are

x + y = 14 …(i)

x – y = 4 …(ii)

From eqn (ii), x = 4 + y …(iii)

On substituting x = 4 + y in eqn (i), we get

⇒ 4 + y + y = 14

⇒ 4 + 2y = 14

⇒ 2y = 14 – 4

⇒ 2y = 10

Now, on putting  in eqn (iii), we get

⇒ x = 4 + 5

⇒ x = 9

Thus, x = 9 andy = 5 is the required solution.

Question 4.

Solve the following pair of linear equations by substitution method:

0.5x + 0.8y = 3.4

0.6x — 0.3y = 0.3

Given equations are

0.5x + 0.8y = 3.4 …(i)

0.6x – 0.3y = 0.3 …(ii)

From eqn (ii), 2x – y = 1

y = 2x – 1 …(iii)

On substituting y = 0.2x – 1 in eqn (i), we get

⇒ 0.5x + 0.8(2x – 1) = 3.4

⇒ 0.5x + 1.6x – 0.8 = 3.4

⇒ 2.1x = 3.4 + 0.8

⇒ 2.1x = 4.2

Now, on putting x = 2 in eqn (iii), we get

⇒ y = 2(2) – 1

⇒ y = 4 – 1

⇒ y = 3

Thus, x = 2 and y = 3 is the required solution.

Question 5.

Solve the following pair of linear equations by substitution method:

x + y = a—b

ax —by = a2 + b2

Given equations are

x + y = a – b …(i)

ax – by = a2 + b2 …(ii)

From eqn (i), x = a – b – y …(iii)

On substituting x = a – b – y in eqn (ii), we get

⇒ a(a – b – y) – by = a2 + b2

⇒ a2 – ab – ay – by = a2 + b2

⇒ – ab – y(a + b) = b2

⇒ – y(a + b) = b2 + ab

Now, on putting y = – b in eqn (iii), we get

⇒ x = a – b – ( – b)

⇒ x = a

Thus, x = a and y = – b is the required solution.

Question 6.

Solve the following pair of linear equations by substitution method:

x + y = 2m

mx — ny = m2 + n2

Given equations are

x + y = 2m …(i)

mx – ny = m2 + n2 …(ii)

From eqn (i), x = 2m – y …(iii)

On substituting x = 2m – y in eqn (ii), we get

⇒ m(2m – y) – ny = m2 + n2

⇒ 2m2 – my – ny = m2 + n2

⇒ – y(m + n) = m2 – 2m2 + n2

⇒ – y(m + n) = – m2 + n2

⇒ y = – (n – m) = m – n

Now, on putting y = m – n in eqn (iii), we get

⇒ x = 2m – (m – n)

⇒ x = 2m – m + n

⇒ x = m + n

Thus, x = m + n and y = m – n is the required solution.

Question 7.

Solve the following system of equations by substitution method:

Given equations are

…(i)

…(ii)

eqn (i) can be re – written as,

⇒ x + 2y = 0.8×2

⇒ x + 2y = 1.6 …(iii)

On substituting x = 1.6 – 2y in eqn (ii), we get

Now, putting the y = 0.6 in eqn (iii), we get

⇒ x + 2y = 1.6

⇒ x + 2(0.6) = 1.6

⇒ x + 1.2 = 1.6

x = 0.4

Thus, x = 0.4 and y = 0.6 is the required solution.

Question 8.

Solve the following system of equations by substitution method:

s —t = 3

Given equations are

s – t = 3 …(i)

…(ii)

From eqn (i), we get

⇒ s = 3 + t …(iii)

On substituting s = 3 + t in eqn (ii), we get

⇒ 6 + 2t + 3t = 6×6

⇒ 5t = 36 – 6

Now, putting the t = 6 in eqn (iii), we get

⇒ s = 3 + 6

s = 9

Thus, s = 9 andt = 6 is the required solution.

Question 9.

Solve the following system of equations by substitution method:

, a ≠ 0, b ≠ 0

ax – by = a2 – b2

Given equations are

…(i)

ax – by = a2 – b2 …(ii)

eqn (i) can be re – written as,

⇒ bx + ay = ab×2

⇒ bx + ay = 2ab

…(iii)

On substituting  in eqn (ii), we get

⇒ 2a2 b – a2 y – b2 y = b(a2 – b2)

⇒ 2a2 b – y(a2 + b2 ) = a2b – b3

– y(a2 + b2 ) = a2 b – b3 – 2a2b

– y(a2 + b2 ) = – a2b – b3

– y(a2 + b2 ) = – b(a2 + b2)

Now, on putting y = b in eqn (iii), we get

⇒ x = a

Thus, x = a and y = b is the required solution.

Question 10.

Solve the following system of equations by substitution method:

x + y = 2ab

Given equations are

…(i)

x + y = 2ab …(ii)

eqn (i) can be re - written as,

⇒ b2 x + a2 y = ab × (a2 + b2)

…(iii)

Now, from eqn (ii), y = 2ab – x …(iv)

On substituting y = 2ab – x in eqn (iii), we get

⇒ b2 x = b3 a – a3b + a2x

⇒ b2x – a2x = b3a – a3b

⇒ (b2 – a2) x = ab(b2 – a2)

x = ab

Now, on putting x = ab in eqn (iv), we get

⇒ y = 2ab – ab

⇒ y = ab

Thus, x = ab and y = ab is the required solution.