Solve the Following Quadratic Equation by Factorization

Question:

\[ \frac{3(3x-1)}{2x+3} – \frac{2(2x+3)}{3x-1} =5, \qquad x\ne\frac13,-\frac32 \]

Solution

Given,

\[ \frac{3(3x-1)}{2x+3} – \frac{2(2x+3)}{3x-1} =5 \]

Multiplying both sides by \((2x+3)(3x-1)\), we get

\[ 3(3x-1)^2-2(2x+3)^2 = 5(2x+3)(3x-1) \] \[ 3(9x^2-6x+1)-2(4x^2+12x+9) = 5(6x^2+7x-3) \] \[ 27x^2-18x+3-8x^2-24x-18 = 30x^2+35x-15 \] \[ 19x^2-42x-15 = 30x^2+35x-15 \] \[ 11x^2+77x=0 \] \[ 11x(x+7)=0 \]

Therefore,

\[ x=0 \] or \[ x=-7 \]

Both values satisfy the condition \(x\ne\frac13,-\frac32\).

Final Answer

\[ \boxed{x=0 \quad \text{or} \quad x=-7} \]