Question

Let $x$ be a value such that $8x^2 + 7x - 1 = 0$ and $24x^2+53x-7 = 0.$ What is the value of $x$? Express your answer as a simplified common fraction.

Answer 1

$8x^2+7x-1=0\ \wedge\ 24x^2+53x-7=0\\\\\text{The equation:}\\\\24x^2+53x-7=8x^2+7x-1\qquad\text{subtract}\ 8x^2\ \text{and}\ 7x\ \text{from both sides}\\\\16x^2+46x-7=-1\qquad\text{add 1 to both sides}\\\\16x^2+46x-6=0\qquad\text{divide both sides by 2}\\\\8x^2+23x-3=0\\\\8x^2+24x-x-3=0\\\\8x(x+3)-1(x+3)=0\\\\(x+3)(8x-1)=0\iff x+3=0\ \vee\ 8x-1=0\\\\x+3=0\qquad\text{subtract 3 from both sides}\\\boxed{x=-3}\\\\8x-1=0\qquad\text{add 1 to both sides}\\8x=1\qquad\text{divide both sides by 8}\\\boxed{x=\dfrac{1}{8}}$