# How do you find the solution of the system of equations 5x - 2y = 4  and 3x + y = 9 ?

May 5, 2018

Find the value of one variable, and use that to solve for the other: See below.

#### Explanation:

A system of equations can technically be solve via one of 2 methods: elimination (I won't explain that here, since it is impractical, and I don't know that one too well; it makes your life necessarily difficult); substitution, where you isolate one of the variables in either equation (let's say you isolate $x$), and replace $x$ in the other equation with the expression that you determined equals $x$. Graphing is good if you're in a pinch, but useless if you don't have a graphing calculator!

I'll demonstrate below:

Substitution (personal recommendation)

Step 1: Isolate one variable in one of the equations

$5 x - 4 y = 4$
$3 x + y = 9$

$5 x \cancel{- 4 y} \textcolor{b l u e}{+ 4 y} = 9 \textcolor{b l u e}{+ 4 y}$

$5 x \textcolor{g r e e n}{- 9} = \cancel{9} + 4 y \textcolor{g r e e n}{- 9}$

$\frac{5}{4} x - \frac{9}{4} = \textcolor{red}{\frac{\cancel{4}}{4}} y$

$\frac{5}{4} x - \frac{9}{4} = y$

Now replace $y$ in the other equation with $\frac{5}{4} x - \frac{9}{4}$, and solve for $x$

$3 x + \frac{5}{4} x - \frac{9}{4} = 9$

$\frac{12}{4} x + \frac{5}{4} x - \frac{9}{4} = 9$

$\frac{17}{4} x = \frac{36}{4} + \frac{9}{4}$

$\frac{17}{4} x = \frac{45}{4}$

$x = \frac{45}{4} \div \frac{17}{4}$

$x = \frac{45 \times 4}{4 \times 17}$
$x = \frac{45 \times \cancel{4}}{\cancel{4} \times 17}$

$x = \frac{45}{17}$

And solve for $y$

$3 \left(\frac{45}{17}\right) - 9 = y$

$x = \frac{45}{17}$ and $y = - \frac{18}{17}$

Hope that helps!