How do you solve $\frac{x + 1}{2} + 1 = \frac{x - 5}{3}$ $(x+1) / 2 + 1 = (x-5) / 3$ ?

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How do you solve $\frac{x + 1}{2} + 1 = \frac{x - 5}{3}$ $(x+1) / 2 + 1 = (x-5) / 3$ ?

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Answer 1 · 2016-07-23T17:26:19

$x = - 19$ $x=-19$

Explanation:

First, we have $\frac{x + 1}{2} + 1 = \frac{x - 5}{3}$ $(x+1)/2 +1 = (x-5)/3$
Simplifying the left hand side, we have,
$\frac{x + 1}{2} + 1$ $(x+1)/2+1$
$\frac{x + 1 + 2}{2}$ $(x+1+2)/2$
$\frac{x + 3}{2}$ $(x+3)/2$
So we now have,
$\frac{x + 3}{2} = \frac{x - 5}{3}$ $(x+3)/2 = (x-5)/3$
Multiply each side with the divisor from the other side.
$(x + 3) \cdot 3 = (x - 5) \cdot 2$ $(x+3)*3=(x-5)*2$
$3 x + 9 = 2 x - 10$ $3x+9=2x-10$
Subtract $2 x$ $2x$ from both sides
$3 x - 2 x + 9 = 2 x - 2 x - 10$ $3x-2x+9 = 2x-2x-10$
$x + 9 = - 10$ $x+9 = -10$
Subtract $9$ $9$ from both sides
$x + 9 - 9 = - 10 - 9$ $x+9-9 = -10-9$
$x = - 19$ $x =-19$

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How do you solve $\frac{x + 1}{2} + 1 = \frac{x - 5}{3}$ $(x+1) / 2 + 1 = (x-5) / 3$ ?

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How do you solve $\frac{x + 1}{2} + 1 = \frac{x - 5}{3}$ $(x+1) / 2 + 1 = (x-5) / 3$ ?