How do you solve (7x)/5+5/2=x/10+1/5 and check for extraneous solutions?

Jan 30, 2017

$x = - \frac{23}{13}$

Explanation:

To eliminate the fractions in this equation, multiply ALL terms on both sides by the $\textcolor{b l u e}{\text{lowest common multiple}}$ (LCM) of the denominators, 2 , 5 and 10

The LCM of 2 ,5 and 10 is 10

$\left({\cancel{10}}^{2} \times \frac{7 x}{\cancel{5}} ^ 1\right) + \left({\cancel{10}}^{5} \times \frac{5}{\cancel{2}} ^ 1\right) = \left({\cancel{10}}^{1} \times \frac{x}{\cancel{10}} ^ 1\right) + \left({\cancel{10}}^{2} \times \frac{1}{\cancel{5}} ^ 1\right)$

$\Rightarrow 14 x + 25 = x + 2 \leftarrow \textcolor{red}{\text{ no fractions}}$

subtract x from both sides.

$14 x - x + 25 = \cancel{x} \cancel{- x} + 2$

$\Rightarrow 13 x + 25 = 2$

subtract 25 from both sides.

$13 x \cancel{+ 25} \cancel{- 25} = 2 - 25$

$\Rightarrow 13 x = - 23$

To solve for x, divide both sides by 13

$\frac{\cancel{13} x}{\cancel{13}} = \frac{- 23}{13}$

$\Rightarrow x = - \frac{23}{13}$

$\textcolor{b l u e}{\text{As a check}}$

Substitute this value into both sides of the equation and if the left side is equal to the right side then it is the solution.

$\left(\frac{7}{5} \times \left(- \frac{23}{13}\right)\right) + \frac{5}{2} = - \frac{161}{65} + \frac{5}{2} = \frac{3}{130}$

$- \frac{23}{130} + \frac{26}{130} = \frac{3}{130}$

left side = right side

$\Rightarrow x = - \frac{23}{13} \text{ is the solution}$

Extraneous solutions are solutions generated by the solving of the equation which do not satisfy the original equation. There are no extra solutions here.