# How do you solve the system x+2y=13 and 3x-5y=6 using substitution?

Mar 17, 2018

$x = 7$
$y = 3$

#### Explanation:

So you can pick either one of the 2 equations listed to get 1 term alone. In explanation I will be using the second one $\left(3 x - 5 y = 6\right)$

So we have $3 x - 5 y = 6$

First we add $5 y$ to both sides giving us
$3 x = 5 y + 6$

Next we divide $3$ on both sides to get $x$ alone, which gives us
$x = \frac{5}{3} y + \frac{6}{3}$

$x = \frac{5}{3} y + 2$

Now that have $x$ we can plug this into the first equation, which is $\left(x + 2 y = 13\right)$

$\frac{5}{3} y + 2 + 2 y = 13$

Now we subtract the $2$ on both sides giving us
$\frac{5}{3} y + 2 y = 11$

Now we add the $y$ values by finding common denominators
$\frac{5}{3} y + \text{(3)2"/"(3)1} y = 11$

Then add the $y$ values
$\frac{5}{3} y + \frac{6}{3} y = 11$

$\frac{5 y + 6 y}{3} = 11$

$\frac{11 y}{3} = 11$

Now we multiply by $3$ on both sides
$11 y = 33$

Next divide by $11$
$y = 3$

So now that we have the value for $y$ we can find $x$ easily by plugging this into any equation that has $y$. For this part I'm going to use the first equation which was, $x + 2 y = 13$

$y = 3$
$x + 2 y = 13$

$x + 2 \left(3\right) = 13$

$x + 6 = 13$

Subtracting 6 from both sides gives us
$x = 7$

There fore
$x = 7$
$y = 3$