# How do you solve  7x – 7 = 2x + 8?

Mar 30, 2016

$x = 3$

#### Explanation:

This is solved by simply moving the terms with $x$ to the same side and the integers to the same side. You can do this by subtracting $2 x$ from both sides and adding $7$ to both sides. This gives:

$5 x = 15$

Then divide both sides by $5$, which gives $x = 3$.

Mar 30, 2016

$x = 3$

#### Explanation:

The one rule you must follow is that what you do to one side of the equation you do to the other. Otherwise the 'equals' sign becomes false

Example: $7 = 7$ this is true as the value on both sides is the same.

If I apply $7 - 2 = 7$ this becomes false as we have 5=7 which is not true. How ever

$7 - 2 = 7 - 2$ is true. We have applied the action to both sides.
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This is why what I wrote is important:

Given:$\text{ } \textcolor{b r o w n}{7 x - 7 = 2 x + 8}$

Subtract $\textcolor{b l u e}{2 x}$ from both sides

$\text{ } \textcolor{b r o w n}{7 x \textcolor{b l u e}{- 2 x} - 7 = 2 x \textcolor{b l u e}{- 2 x} + 8}$

But $2 x - 2 x = 0$

$\text{ } 5 x - 7 = 0 + 8$

Add $\textcolor{b l u e}{7}$ to both sides

" "color(brown)(5x-7color(blue)(+7)=8color(blue)(+7)

But $- 7 + 7 = 0$

$\text{ } 5 x + 0 = 15$

Divide both sides by $\textcolor{b l u e}{5}$

$\text{ } \textcolor{b r o w n}{\frac{5}{\textcolor{b l u e}{5}} x = \frac{15}{\textcolor{b l u e}{5}}}$

But $\frac{5}{5} = 1 \text{ and } \frac{15}{5} = 3$

$\text{ } \textcolor{m a \ge n t a}{x = 3}$

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