Question

For solving pair of equation, in this exercise use the method of elimination by equating coefficients :
2x − 3y − 3 = 0
`[2x]/3 + 4y + 1/2` = 0

Answer 1

2x − 3y − 3 = 0
⇒ 2x - 3y = 3          ...(1)

`[2x]/3 + 4y + 1/2` = 0
Multiply by 6,
`6 xx [2x]/3 + 6 xx 4y + 1/2 xx 6 = 0 xx 6`
4x + 24y = - 3          ...(2)

Multiplying equation no. (1) by 8
16x - 24y = 24           ...(3)

Adding equation (3) and (2)
         16x - 24y = 24
      +  4x + 24y = - 3 
                  20x = 21
                      x = `21/20`

From (1)
∴ `2[21/20]` - 3y = 3
∴ - 3y = 3 - `21/20`
∴ y = `-3/10`

Answer 2

1. 2x − 3y − 3 = 0
2. `(2x)/3 + 4y + 1/2 = 0`

Multiply the entire second equation by 6 to eliminate the denominators:

`6 ((2x)/3 + 4y + 1/2) = 0`

4x + 24y + 3 = 0

Write both equations in standard form

Equation (1): 2x − 3y = 3

Equation (2): 4x + 24y = −3

Let’s eliminate x. Multiply Equation (1) by 2:

(2x − 3y = 3) × 2 ⇒ 4x − 6y = 6

(4x − 6y) − (4x + 24y) = 6 − (−3)

4x − 6y − 4x − 24y = 6 + 3

`-30y = 9 => y = (-9)/30 = (-3)/10`

`2x - 3 (-3/10) = 3 => 2x + 9/10 = 3 => 2x = 3 -9/10 = (30-9)/10 = 21/10 => x=21/20`

`x = 21/20,  y = (-3)/10`