How do you solve abs(7-2x)=5?

Aug 26, 2017

$x = 1 \mathmr{and} 6$

Explanation:

The use of the special brackets of | | means that whatever is inside them is considered as positive.

On the right side of the equals we have +5.

So the left side must end up as $| \pm 5 |$ giving:

$| \pm 5 | = + 5$

Ok!

Lets consider what will turn $7 - 2 x$ into -5

Set $\textcolor{w h i t e}{\text{d}} 7 - 2 x = - 5$

$2 x = 7 + 5$

$x = \frac{12}{2} = + 6$

Check $\left(\textcolor{w h i t e}{\frac{2}{2}} 7 - \left[2 \times 6\right] \textcolor{w h i t e}{\text{d}}\right) \to 7 - 12 = - 5$ as required

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Lets consider what will turn $7 - 2 x$ into +5

Set $\textcolor{w h i t e}{\text{d}} 7 - 2 x = + 5$

$2 x = 7 - 5$

$x = \frac{2}{2} = 1$

Check $\left(\textcolor{w h i t e}{\frac{2}{2}} 7 - \left[2 \times 1\right] \textcolor{w h i t e}{\text{d}}\right) \to 5 \times 1 = + 5$ as required

Aug 26, 2017

See a solution process below:

Explanation:

The absolute value function takes any term and transforms it to its non-negative form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent.

Solution 1

$7 - 2 x = - 5$

$- \textcolor{red}{7} + 7 - 2 x = - \textcolor{red}{7} - 5$

$0 - 2 x = - 12$

$- 2 x = - 12$

$\frac{- 2 x}{\textcolor{red}{- 2}} = \frac{- 12}{\textcolor{red}{- 2}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 2}}} x}{\cancel{\textcolor{red}{- 2}}} = 6$

$x = 6$

Solution 2

$7 - 2 x = 5$

$- \textcolor{red}{7} + 7 - 2 x = - \textcolor{red}{7} + 5$

$0 - 2 x = - 2$

$- 2 x = - 2$

$\frac{- 2 x}{\textcolor{red}{- 2}} = \frac{- 2}{\textcolor{red}{- 2}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 2}}} x}{\cancel{\textcolor{red}{- 2}}} = 1$

$x = 1$

The Solutions Are: $x = 6$ and $x = 1$