# How do you solve the following system?:  1/2x=y-1 , 1/2x+1/3y=5

Mar 28, 2016

$\textcolor{b l u e}{x = 7}$

$\textcolor{b l u e}{y = \frac{9}{2}}$

#### Explanation:

Notice that both equations coefficient of $x \text{ is } + \frac{1}{2}$

This makes life easy as we can 'get rid' of one for the variables.

To solve an equation easily you need just one variable and a relationship to some constant.
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Given:
$\frac{1}{2} x = y - 1$............................(1)
$\frac{1}{2} x + \frac{1}{3} y = 5$......................(2)

It is common practice to convert them so that they are of form
$y = m x + c$. This not strictly necessary in this case as the process will sort this out any way.

Equation (2) has the lesser value of $y$ so I am opting for:

$\textcolor{b r o w n}{\text{Subtract equation (2) from equation (1) giving:}}$

$\textcolor{g r e e n}{\text{Notice how I have used 0 as a place keeper}}$

$\text{ } \frac{1}{2} x + 0 y \textcolor{w h i t e}{.} = y - 1$
$\text{ "underline(1/2x+1/3y=0y+5 )->"subtract}$
" "color(green)(0" " -1/3y=y-6)

Add $\textcolor{m a \ge n t a}{\frac{1}{3} y}$ to both sides

$\text{ } \textcolor{g r e e n}{- \frac{1}{3} y \textcolor{w h i t e}{.} \textcolor{m a \ge n t a}{+ \frac{1}{3} y} = y \textcolor{w h i t e}{.} \textcolor{m a \ge n t a}{+ \frac{1}{3} y} - 6}$

$\text{ } 0 = \frac{4}{3} y - 6$

Add 6 to both sides giving:

$6 = \frac{4}{3} y$

$\textcolor{b l u e}{y = \frac{6 \times 3}{4} = \frac{9}{2}}$
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I am choosing equation (1) for substitution of y

$\frac{1}{2} x = y - 1 \text{ "->" } \frac{1}{2} x = \frac{9}{2} - 1$

$\textcolor{b l u e}{x = 7}$
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Check

$\frac{1}{2} x + \frac{1}{3} y = 5 \text{ becomes } \frac{1}{2} \left(7\right) + \frac{1}{3} \left(\frac{9}{2}\right) = 5$ Confirmed!