Question

Solve: $\frac{4x-3}{2x+1}-10\left(\frac{2x+1}{4x-3}\right)=3$

• A. $x=\frac{1}{2},x=-\frac{1}{7}$
• B. $x=\frac{6}{7},x=\frac{1}{2}$
• C. $x=-\frac{4}{3},x=\frac{1}{8}$
• D. $x=-\frac{9}{2},x=-\frac{1}{7}$

Answer 1

The correct option is C $x=-\frac{4}{3},x=\frac{1}{8}$

$\frac{4x-3}{2x+1}-10\left(\frac{2x+1}{4x-3}\right)=3$

$\Rightarrow (4x-3{)}^2-10(2x+1{)}^2=3(2x+1)(4x-3)$

$\Rightarrow 16x^2-24x+9-10(4x^2+4x+1)=3(8x^2+4x-6x-3)$

$\Rightarrow 16x^2-40x+9-40x^2-40x-10=24x^2-6x-9$

$\Rightarrow 16x^2-40x^2-24x^2-24x-40x+6x+9-10+9=0$

$-48x^2-58x+8=0$

$\Rightarrow 24x^2+29x-4=0$

$\Rightarrow 24x^2+32x-3x-4=0$

$\Rightarrow 8x(3x+4)-1(3x+4)=0$

$\Rightarrow (8x-1)(3x+4)=$

$\therefore 8x-1=0$

$\Rightarrow x=\frac{1}{8}$

$\therefore 3x+4=0$

$\Rightarrow x=-\frac{4}{3}$