Solve the Following Quadratic Equation by Factorization
Question:
\[ \frac{x+3}{x+2}=\frac{3x-7}{2x-3}, \qquad x\ne -2,\frac{3}{2} \]Solution
Given:
\[ \frac{x+3}{x+2}=\frac{3x-7}{2x-3} \]Cross-multiplying, we get:
\[ (x+3)(2x-3)=(x+2)(3x-7) \]Expanding both sides:
\[ 2x^2+3x-9=3x^2-x-14 \]Bringing all terms to one side:
\[ x^2-4x-5=0 \]Factorizing:
\[ x^2-5x+x-5=0 \] \[ x(x-5)+1(x-5)=0 \] \[ (x-5)(x+1)=0 \]Therefore,
\[ x-5=0 \quad \text{or} \quad x+1=0 \] \[ x=5 \quad \text{or} \quad x=-1 \]Both values satisfy the condition \(x\ne -2,\frac{3}{2}\).