Question

Solve the following equation by factorisation method:

`2((2x+3)/(x-3))-25((x-3)/(2x+3)) = 5, x\cancel=3, x\cancel= -3/2`

Answer 1

Given:

`2((2x+3)/(x-3))-25((x-3)/(2x+3)) = 5, x\cancel=3, x\cancel= -3/2`

Multiply both sides:

`(x−3)(2x+3)xx [2 ((2x+3)/(x-3))-25((x-3)/(2x+3))]=5(x−3)(2x+3)`

`(x−3)(2x+3)xx2((2x+3)/(x-3)) = 2(2x+3)^2`

`(x−3)(2x+3)xx25 ((x-3)/(2x+3)) = 25(x-3)^2`

5(x − 3) (2x + 3)

2(2x + 3)² − 25 (x − 3)² = 5(x − 3) (2x + 3)

Expand all terms

(2x + 3)² = x² + 12x + 9

⇒ 2(4x² + 12x + 9) = 8x² + 24x + 18

(x − 3)² = x² − 6x + 9

⇒ 25(x² − 6x + 9) = 25x² − 150x + 225

(x − 3) (2x + 3) = 2x² + 3x − 6x − 9 = 2x² − 3x − 9

⇒ 5(2x² − 3x − 9) = 10x² − 15x

(8x² + 24x + 18) − (25x² − 150x + 225) = 10x² − 15x − 45

8x² + 24x + 18 − 25x² + 150x − 225 = −17x² + 174x − 207

−17x² + 174x − 207 = 10x² − 15x − 45

Bring all RHS terms to LHS:

−17x² + 174x − 207 − 10x² + 15x + 45 = 0

−27x² + 189x − 162 = 0

Multiply both sides by −1 to make leading coefficient positive:

27x² − 189x + 162 = 0

First simplify the equation by dividing all terms by 9:

3x² − 21x + 18 = 0

We need two numbers whose product is 3 × 18 = 543 and sum is −21.

Numbers: −3 and −18

3x² − 3x − 18x + 18 = 0

⇒ 3x(x − 1) − 18(x − 1) = 0

⇒ (3x − 18) (x − 1) = 0

3(x − 6) (x − 1) = 0

x = 6 or x = 1