Solve the following quadratic equations by factorization: 3 (3x - 1)/(2x + 3) -2 (2x + 3)/(3x - 1) = 5; x ≠ 1/3, -3/2

Question

Solve the following quadratic equations by factorization:

\(3(\frac{3x-1}{2x+3})-2(\frac{2x+3}{3x-1})=5;x\neq\frac{1}{3},-\frac{3}{2}\)

Answer 1

Given: \(3(\frac{3x-1}{2x+3})-2(\frac{2x+3}{3x-1})=5;x\neq\frac{1}{3},-\frac{3}{2}\)

to find: Solution of the above quadratic equation.

Solution: In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

\(3(\frac{3x-1}{2x+3})-2(\frac{2x+3}{3x-1})=5;x\neq\frac{1}{3},-\frac{3}{2}\)

⇒ 3(9x² + 1 – 6x) – 2(4x² + 9 + 12x) 

= 5(6x² – 3 + 7x)

⇒ 27x² + 3 – 18x - 8x² - 18 – 24x 

= 30x² - 15 + 35x

⇒ 19x² - 42x – 15 

= 30x² - 15 + 35x 

⇒19x² - 30x² - 42x – 35x - 15 + 15 = 0

⇒ -11x2 - 77x = 0

⇒ 11x2 + 77x = 0

⇒ 11x(x + 7) = 0

⇒ x =0  and (x + 7)=0

⇒ x = 0 and x = -7

⇒ x = 0, - 7