# Question 498a4

Sep 1, 2017

$y = - \frac{9}{5}$
$x = 15$

#### Explanation:

The most straight forward way to resolve this particular is to get rid of the fraction as a first step.

$\frac{1}{5} x + y = \frac{6}{5} \text{ } \ldots \ldots \ldots \ldots \ldots . . E q u a t i o n \left(1\right)$
$\frac{1}{10} x + \frac{1}{3} y = \frac{9}{10} \text{ } \ldots \ldots \ldots . E q u a t i o n \left(2\right)$
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$\textcolor{b l u e}{\text{Consider Eqn(1)}}$

$\textcolor{g r e e n}{\left[\frac{x}{5}\right] \textcolor{w h i t e}{\text{d")+color(white)("d")[ycolor(red)(xx1)]color(white)("d}} = \left[\frac{6}{5}\right]}$

$\textcolor{g r e e n}{\left[\frac{x}{5}\right] \textcolor{w h i t e}{\text{d")+color(white)("d")[ycolor(red)(xx5/5)]color(white)("d}} = \left[\frac{6}{5}\right]}$

$\frac{x}{5} + \frac{5 y}{5} = \frac{6}{5}$

Multiply both sides by 5

$x + 5 y = 6 \text{ } \ldots \ldots \ldots \ldots \ldots \ldots \ldots . E q u a t i o n \left({1}_{a}\right)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{b l u e}{\text{Consider Eqn(2)}}$

$\textcolor{g r e e n}{\left[\frac{x}{10} \textcolor{red}{\times 1}\right] + \left[\frac{y}{3} \textcolor{red}{\times 1}\right] = \left[\frac{9}{10} \textcolor{red}{\times 1}\right]}$

$\textcolor{g r e e n}{\left[\frac{x}{10} \textcolor{red}{\times \frac{3}{3}}\right] + \left[\frac{y}{3} \textcolor{red}{\times \frac{10}{10}}\right] = \left[\frac{9}{10} \textcolor{red}{\times \frac{3}{3}}\right]}$

$\frac{3 x}{30} + \frac{10 y}{30} = \frac{27}{30}$

Multiply both sides by 30

$3 x + 10 y = 27 \text{ } \ldots \ldots \ldots . . E q u a t i o n \left({2}_{a}\right)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Putting it all together}}$

$x + 5 y = 6 \text{ } \ldots \ldots \ldots \ldots \ldots \ldots E q n \left({1}_{a}\right)$
$3 x + 10 y = 27 \text{ } \ldots \ldots \ldots \ldots E q n \left({2}_{a}\right)$

$E q n \left({2}_{a}\right) - \left(3 \times E q n \left({1}_{a}\right)\right)$

$3 x + 10 y = 27 \text{ "larr" } E q n \left({2}_{a}\right)$
$\underline{3 x + 15 y = 18} \text{ "larr" } 3 \times E q n \left({1}_{a}\right)$
$0 x - \textcolor{w h i t e}{2} 5 y = 9$

Multiply both sides by (-1)

$+ 5 y = - 9$

Divide both sides by 5

$y = - \frac{9}{5} \text{ } \ldots \ldots \ldots \ldots . E q n \left(3\right)$

I choose $E q n \left({1}_{a}\right)$ it is works out more easily. You would still get the same answer if you used $E q n \left({2}_{a}\right)$ but it would involve more work.

Using $E q n \left(3\right)$ substitute for $y$ in $E q n \left({1}_{a}\right)$

color(green)(x+5color(red)(y)=6" "->" "x+(5color(red)(xx-9/5))=6#

$x - 9 = 6$

$x = 6 + 9 = 15$