- 3x - y = 3 9x — 3y = 9 Solve the following pair of linear equations by…
- 7x— 15y = 2 x + 2y = 3 Solve the following pair of linear equations by…
- x + y = 14 x - y = 4 Solve the following pair of linear equations by…
- 0.5x + 0.8y = 3.4 0.6x — 0.3y = 0.3 Solve the following pair of linear…
- x + y = a—b ax —by = a^2 + b^2 Solve the following pair of linear equations by…
- x + y = 2m mx — ny = m^2 + n^2 Solve the following pair of linear equations by…
- x/2 + y = 0.8 x + y/2 = 7/10 Solve the following system of equations by…
- s —t = 3 s/3 + t/2 = 6 Solve the following system of equations by substitution…
- x/a + y/b = 2 , a ≠ 0, b ≠ 0 ax - by = a^2 - b^2 Solve the following system of…
- 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
Answer:
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
Answer:
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
Answer:
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 and
y = 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
Answer:
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
Answer:
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
Answer:
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:
Answer:
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
Answer:
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
Answer:
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
Answer:
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.
