# How do you solve 1/8x-5=3?

Oct 30, 2015

$x = 64$ See detailed explanation for how you do it!

#### Explanation:

$\textcolor{red}{\text{Change things so that there is only one}}$ $\textcolor{red}{x}$ $\textcolor{red}{\text{and it is on the left of the = sign. Everything else is to be on the right of the = sign}}$

Brackets use only to show what parts are being changed. The brackets are only for demonstration to you what is happening. They serve no other purpose.

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Step1
Move the 5 from left to right

color(brown)("In add or subtract, if we can change a number to zero" $\textcolor{B r o w n}{\text{it has no effect. This is the same as removing it from that side of the equation.}}$

Add 5 to both sides giving
$\left(\frac{1}{8} x - 5\right) + 5 = \left(3\right) + 5$

$\frac{1}{8} x + 0 = 8$ ................................(1)

I have used plus 5 because negative 5 and plus 5 make zero.
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Step 2

Rewrite equation (1) as:

$\frac{1}{8} x = 8$ ...............................(2)

$\textcolor{b r o w n}{\text{In this case if we can change the}}$ $\textcolor{b r o w n}{\frac{1}{8}}$$\textcolor{b r o w n}{\text{into 1 we are then multiplying}}$ $\textcolor{b r o w n}{x}$ $\textcolor{b r o w n}{\text{by 1. Consequently not changing its value}}$

Multiply both sides of equation (2) by 8

$\left(\frac{1}{8} x\right) \times 8 = \left(8\right) \times 8$

But $\frac{1}{8} \times 8$ is the same as $\frac{8}{8}$ which is 1 so now we have changed equation (2) to

$1 \times x = 64$
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There you have it: Only $x$ on one side and its value on the other.