How do you solve $\frac{1}{b + 2} + \frac{1}{b + 2} = \frac{3}{b + 1}$ $1/(b+2)+1/(b+2)=3/(b+1)$ ?

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Answer 1 · 2017-07-13T07:44:07

By arranging equation, you will get $b = - 4$ $b=-4$

When you arrange your equation, you will get

$\frac{1 + 1}{b + 2} = \frac{3}{b + 1}$ $(1+1)/(b+2) = 3/(b+1)$

$\frac{2}{b + 2} = \frac{3}{b + 1}$ $2/(b+2) = 3/(b+1)$

$2 (b + 1) = 3 (b + 2)$ $2(b+1) = 3(b+2)$

$2 b + 2 = 3 b + 6$ $2b + 2 = 3b + 6$

$2 b - 3 b = 6 - 2$ $2b-3b = 6-2$

$- b = 4$ $-b = 4$

or

$b = - 4$ $b=-4$

This is your answer $b = - 4$ $b=-4$

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