How do you solve $\frac{8}{4 x + 5} - \frac{1}{x} = \frac{4}{7 x - 1}$ $\frac { 8} { 4x + 5} - \frac { 1} { x } = \frac { 4} { 7x - 1}$ ?

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Answer 1 · 2017-10-30T16:47:24

follow the procedure below

From LHS,

equate the whole thing to the same denominator :

$\times$ $xx$ by LCM { (4x+5) (x) }

simplifying worked out resulting in:

$\frac{8 x - (4 x + 5)}{(4 x - 5) (x)}$ $(8x -(4x+5))/((4x-5)(x))$ = $\frac{4}{7 x - 1}$ $4/(7x - 1)$

further simplification:

$\frac{8 - 4 x + 5}{4 x^{2} + 5 x}$ $(8-4x+5)/(4x^2 +5x)$ = $\frac{4}{7 x + 1}$ $4/(7x +1)$

carry out cross multiplication to obtain the two values of x

How do you solve 84x+5−1x=47x−1\frac { 8} { 4x + 5} - \frac { 1} { x } = \frac { 4} { 7x - 1}?