Question

Please I need help!!

Solve for x.

Answer 1

triangles are similar by the Angle Bisector Theorem.

`\cfrac{10x-2}{7x}=\cfrac{8x}{5x+3}\implies \stackrel{ F~O~I~L }{(10x-2)(5x+3)}=(7x)(8x) \\\\\\ 50x^2+20x-6=56x^2\implies 20x-6=6x^2\implies 0=6x^2-20x+6 \\\\\\ 0=2(3x^2-10x+3)\implies 0=3x^2-10x+3 \\\\\\ 0=(3x-1)(x-3)\implies x= \begin{cases} (1)/(3)\\\\ 3 \end{cases}`

now, both values are positive and valid, and both are feasible values of "x", so both are good.