Question

Find the solution set of the pair of linear
equations 2x + y = 8 and 3x - 2y = 12
Oy​

Answer 1

GIVEN :-

• 2x + y = 8 .
• 3x - 2y = 12.

TO FIND :-

• The value of x and y.

SOLUTION :-

✭ 2x + y = 8 ....Eq(1)

✭ 3x - 2y = 12. ....Eq(2)

From Equation 1 we have,

➠ 2x + y = 8

➠ 2x = 8 - y

➠ x = (8 - y)/2

Substitute the value of x in Equation 2,

➠ 3x - 2y = 12

➠ 3 × (8 - y)/2 - 2y = 12

➠ 3(8 - y)/2 - 2y = 12

➠ (24 - 3y)/2 - 2y = 12

➠ (24 - 3y)/2 - 4y/2 = 12

➠ (24 - 3y - 4y)/2 = 12

➠ (24 - 7y)/2 = 12

➠ 24 - 7y = 12 × 2

➠ 24 - 7y = 24

➠ -7y = 24 - 24

➠ -7y = 0

➠ y = 0/-7

➠ y = 0

Substitute the value of y in Equation 1,

➠ 2x + y = 8

➠ 2x + 0 = 8

➠ 2x = 8 - 0

➠ 2x = 8

➠ x = 8/2

➠ x = 4

Hence the required value of x is 4 and the required value of y is 0.

Answer 2

Answer:

⏺️Question

Find the solution set of the pair of linear equations 2x + y = 8 and 3x - 2y = 12

⏺️To find

Solution of the given equation set.

⏺️ Solution

Equation 1 - 2x + y = 8

Equation 2 - 3x - 2y = 12

Now from equation 1 we have

$2x + y = 8$

$2x = 8 - y$

$x = \frac{8 - y}{2}$.

[Substituting value of equation 2]

$3x - 2y = 12$

$3 \times \frac {(8 - y)}{2 - 2y} = 12$

$3 \frac {8-y}{2-2y}$

$\frac{24 - 3y}{2 - 2y} = 12$

$\frac{24 - 3y}{2} - \frac{4y}{2} = 12$

$\frac {24- 3y - 4y}{2} = 12$

$\frac {24 - 7y}{2} = 12$

$24 - 7 y = 2×12$

$24 - 7y = 24$

$-7y = 24 - 24$

$-7y = 0$

$y = \frac{0}{ - 7}$

$y = 0$

[Substituting value of equation 1]

$2x + y = 8$

$2x + 0 = 8$

$2x = 8$

$x = \frac{8}{2}$

$x = 4$

Hence the required value of x is 4 and y is 0

Verification

Equation 1

$2x + y = 8$

$2 \times 4 + 0 = 8$

$8 = 8$

LHS = RHS

Equation 2

$3x - 2y = 12$

$3 \times 4 - 2 \times 0 = 12$

$12 - 0 = 12$

$12 = 12$

LHS = RHS

Hence proved