# How do you solve (x+2)/6 + (x-2)/10 = 6/5?

May 23, 2017

$x = 4$

#### Explanation:

$\textcolor{b l u e}{\text{3 Key points}}$

A fraction's structure consists of :

$\left(\text{count")/("size indicator of what you are counting") ->("numerator")/("denominator}\right)$

You can NOT DIRECTLY add or subtract the counts unless the size indicators are the same.

Multiply by 1 and you do not change the intrinsic value. However, 1 comes in many forms so you can change the way a fraction looks without changing its intrinsic value.
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$\textcolor{b l u e}{\text{Method: }}$

Make all the denominators the same but insuring that the numerators remain proportional to the denominator.

Then just solve for the numerators only. This really works!

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$\textcolor{b l u e}{\text{Answering the question}}$

I choose to make the denominators 60:

color(green)([(x+2)/6color(red)(xx1)] +[(x-2)/10color(red)(xx1)] = [6/5color(red)(xx1)]

color(green)([(x+2)/6color(red)(xx10/10)] +[(x-2)/10color(red)(xx6/6)] = [6/5color(red)(xx12/12)]

color(white)(.)color(green)([(10x+20)/60] color(white)(..)+color(white)(.)[(6x-12)/60]=[72/60]

Thus it also true that:

$\text{ "10x+20" "+" "6x-12" } = 72$

$16 x + 8 = 72$

Subtract 8 from both sides

$16 x = 64$

Divide both sides by 16

$x = \frac{64}{16} = 4$

May 23, 2017

$x = 4$

#### Explanation:

If you have fractions which are in an equation, you can get rid of them immediately.

Multiply each term by the LCM of the denominators. (In this case $\textcolor{b l u e}{30}$)

If you multiply the whole equation by $30$ you do not change the value.

$\frac{\textcolor{b l u e}{30} \times \left(x + 2\right)}{6} + \frac{\textcolor{b l u e}{30} \times \left(x - 2\right)}{10} = \frac{\textcolor{b l u e}{30 \times} 6}{5}$

Now cancel each denominator.

$\frac{{\textcolor{b l u e}{\cancel{30}}}^{5} \times \left(x + 2\right)}{\cancel{6}} + \frac{{\textcolor{b l u e}{\cancel{30}}}^{3} \times \left(x - 2\right)}{\cancel{10}} = \frac{\textcolor{b l u e}{{\cancel{30}}^{6} \times} 6}{\cancel{5}}$
$5 \left(x + 2\right) + 3 \left(x - 2\right) = 6 \times 6 \text{ } \leftarrow$ no more fractions!

$5 x + 10 + 3 x - 6 = 36 \text{ } \leftarrow$ now solve for $x$

$8 x = 36 - 4$

$8 x = 32$

$x = 4$