# How do you solve 3\frac { 11} { 24} - x = 1 1/6+ 1\frac { 1} { 9}?

Aug 2, 2017

$x = \frac{85}{72} = 1 \frac{13}{72}$

#### Explanation:

$3 \frac{11}{24} - x = 1 \frac{1}{6} + 1 \frac{1}{9}$

Convert all the mixed fractions into improper fractions. To do that we multiply the whole number with the denominator, add the product to the numerator, and place the result on the same denominator.

$\frac{83}{24} - x = \frac{7}{6} + \frac{10}{9}$

Now multiply all terms by the LCM of the denominator. We calculate the LCM by writing down the multiples of each denominator.

$24 \implies 24 , 48 , \textcolor{b l u e}{72}$
$6 \implies 6 , 12 , 18 , 24 , 30 , 36 , 42 , 48 , 54 , 60 , 66 , \textcolor{b l u e}{72}$
$9 \implies 9 , 18 , 27 , 36 , 45 , 54 , 63 , \textcolor{b l u e}{72}$

Multiplying all terms by $72$, we write.

$\left(72 \times \frac{83}{24}\right) - \left(72 x\right) = \left(72 \times \frac{7}{6}\right) + \left(72 \times \frac{10}{9}\right)$

$\left(3 \cancel{72} \times \frac{83}{\cancel{24}}\right) - \left(72 x\right) = \left(12 \cancel{72} \times \frac{7}{\cancel{6}}\right) + \left(8 \cancel{72} \times \frac{10}{\cancel{9}}\right)$

$\left(3 \times 83\right) - 72 x = \left(12 \times 7\right) + \left(8 \times 10\right)$

$249 - 72 x = 84 + 80$

$249 - 72 x = 164$

Subtract $164$ from each side.

$249 - 164 - 72 x = 164 - 164$

$85 - 72 x = 0$

Add $72 x$ to each side.

$85 = 72 x$

Divide both sides by $72$.

$\frac{85}{72} = x$

$x = \frac{85}{72} = 1 \frac{13}{72}$