Question

How do I do this? And what’s the answer?

Answer 1

`\cfrac{1}{2x-3}~~ + ~~\cfrac{1}{4x+9}~~ = ~~7 \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{(2x-3)(4x+9)}}{(2x-3)(4x+9)\left( \cfrac{1}{2x-3}~~ + ~~\cfrac{1}{4x+9} \right)}~~ = ~~(2x-3)(4x+9)(7) \\\\\\ (4x+9)~~ + ~~(2x-3)=\stackrel{ \textit{F ~~ O ~~ I ~~ L} }{(2x-3)(28x+63)} \\\\\\ 6x+6~~ = ~~56x^2+126x-84x-189\implies 6x+6~~ = ~~56x^2+42x-189`

`7(8x^2+6x-27)=0\implies 8x^2+6x-27=0\implies (2x-3)(4x+9)=0 \\\\[-0.35em] ~\dotfill\\\\ 2x-3=0\implies 2x=3\implies x=\cfrac{3}{2}\implies \boxed{x=1.5} \\\\[-0.35em] ~\dotfill\\\\ 4x+9=0\implies 4x=-9\implies x=-\cfrac{9}{4}\implies \boxed{x\approx -2.3}`