Solve the Following Quadratic Equation by Factorization
Question:
\[ \frac{3(7x+1)}{5x-3} – \frac{4(5x-3)}{7x+1} = 11, \qquad x\ne\frac35,-\frac17 \]Solution
Given,
\[ \frac{3(7x+1)}{5x-3} – \frac{4(5x-3)}{7x+1} = 11 \]Multiplying both sides by \((5x-3)(7x+1)\), we get
\[ 3(7x+1)^2 – 4(5x-3)^2 = 11(5x-3)(7x+1) \] \[ 3(49x^2+14x+1) – 4(25x^2-30x+9) = 11(35x^2-16x-3) \] \[ 147x^2+42x+3 – 100x^2+120x-36 = 385x^2-176x-33 \] \[ 47x^2+162x-33 = 385x^2-176x-33 \] \[ 338x^2-338x=0 \] \[ 338x(x-1)=0 \]Therefore,
\[ x=0 \] or \[ x=1 \]Both values satisfy the condition \(x\ne\frac35,-\frac17\).