What is the solution? $\frac{x + 5}{5} = 1 - (\frac{x + 5}{7})$ $(x+5)/5=1-((x+5)/7)$

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Answer 1 · 2018-06-06T15:55:26

The solution is $x = - \frac{25}{12}$ $x = -25/12$

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

Multiply both sides by 35:

$7 (x + 5) = 35 - 5 (x + 5)$ $7(x+5)=35-5(x+5)$

Distribute the 7 and the -5:

$7 x + 35 = 35 - 5 x - 25$ $7x+35=35-5x-25$

Add $5 x - 35$ $5x-35$ to both sides:

$12 x = - 25$ $12x=-25$

$x = - \frac{25}{12}$ $x = -25/12$

Answer 2 · 2018-06-06T15:56:25

$\to \frac{x + 5}{5} = 1 - \frac{x + 5}{7}$ $rarr(x+5)/5=1-(x+5)/7$

$\to \frac{x + 5}{5} + \frac{x + 5}{7} = 1$ $rarr(x+5)/5+(x+5)/7=1$

$\to (x + 5) [\frac{1}{5} + \frac{1}{7}] = 1$ $rarr(x+5)[1/5+1/7]=1$

$\to (x + 5) [\frac{12}{35}] = 1$ $rarr(x+5)[12/35]=1$

$\to x = \frac{35}{12} - 5 = - \frac{25}{12}$ $rarrx=35/12-5=-25/12$

What is the solution? x+55=1−(x+57)(x+5)/5=1-((x+5)/7)