# How do you solve the system of equations 2x + 3y = 18 and 3x - 5= y?

Nov 9, 2017

See a solution process below:

#### Explanation:

Step 1) Because the second equation is already solved for $y$, we can substitute $\left(3 x - 5\right)$ for $y$ in the first equation and solve for $x$:

$2 x + 3 y = 18$ becomes:

$2 x + 3 \left(3 x - 5\right) = 18$

$2 x + \left(3 \times 3 x\right) - \left(3 \times 5\right) = 18$

$2 x + 9 x - 15 = 18$

$\left(2 + 9\right) x - 15 = 18$

$11 x - 15 = 18$

$11 x - 15 + \textcolor{red}{15} = 18 + \textcolor{red}{15}$

$11 x - 0 = 33$

$11 x = 33$

$\frac{11 x}{\textcolor{r e}{11}} = \frac{33}{\textcolor{r e}{11}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{11}}} x}{\cancel{\textcolor{r e}{11}}} = 3$

$x = 3$

Step 2) Substitute $3$ for $x$ in the second equation and solve for $y$:

$3 x - 5 = y$ becomes:

$\left(3 \times 3\right) - 5 = y$

$9 - 5 = y$

$4 = y$

$y = 4$

The Solution Is: $x = 3$ and $y = 4$ or $\left(3 , 4\right)$