# How do you solve the system x/2+(2y)/3=32 and x/4-(5y)/9=40?

Feb 22, 2017

$x = 100 \mathmr{and} y = - 27$

#### Explanation:

At first glance this horrifying because of the fractions!

Luckily with equations you can always get rid of any fractions by multiplying each term by the LCM of the denominators.

$\times 6 \rightarrow \text{ "x/2+(2y)/3=32" } \rightarrow 3 x + 4 y = 192$

$\times 36 \rightarrow \text{ "x/4-(5y)/9=40" } \rightarrow 9 x - 20 y = 1440$

The equations look much better, now we can solve them:

$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots . .} 3 x + 4 y = 192. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . A$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots . .} 9 x - 20 y = 1440. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . B$

$A \times - 3 : \text{ } \textcolor{b l u e}{- 9 x} - 12 y = - 576. \ldots \ldots \ldots \ldots \ldots \ldots \ldots C$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} \textcolor{b l u e}{9 x} - 20 y = 1440. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots B$

$C + B : \text{ } - 32 y = 864$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .} \textcolor{red}{y = - 27}$

Substitute $- 27$ for $y$ in $A$

$\text{ } 3 x + 4 \left(- 27\right) = 192. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . A$
$\text{ } 3 x - 108 = 192$
$\text{ } 3 x = 192 + 108$
$\text{ } 3 x = 300$
$\text{ } \textcolor{b l u e}{x = 100}$

Check the solutions in equation B

" Is " 9(100)-20(-27)= 1440 ?
$\text{ } 900 + 540 = 1440$
$\text{ Indeed! } 1440 = 1440$

We can even check in the original equation for A;

$\frac{x}{2} + \frac{2 y}{3} = 32$

$\frac{100}{2} + \frac{2 \times - 27}{3}$

$= 50 - 18$

$= 32 \text{ } \leftarrow$ the answer is correct.