How do you solve $\frac{7 x}{x - 5} = \frac{x + 3}{x}$ $\frac{7x}{x-5}=\frac{x+3}{x}$ ?

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Answer 1 · 2018-03-20T08:33:33

I personally believe you can't

multiply by the denominators to the opposite sides to make everything on one level:

$x (7 x) = x - 5 (x + 3)$ $x(7x)=x-5(x+3)$

then multiply out the brackets@

$7 x^{2} = x^{2} - 2 x - 15$ $7x^2=x^2-2x-15$

then bring everything over on one side:

$6 x^{2} + 2 x + 15 = 0$ $6x^2+2x+15=0$

and then you can't factorise...

How do you solve 7xx−5=x+3x\frac{7x}{x-5}=\frac{x+3}{x}?