# How do you solve the following system:  3x – 5y = 53 , 4x+y=10 ?

May 1, 2017

See the entire solution process below:

#### Explanation:

Step 1) Solve the second equation for $y$:

$4 x + y = 10$

$- \textcolor{red}{4 x} + 4 x + y = - \textcolor{red}{4 x} + 10$

$0 + y = - 4 x + 10$

$y = - 4 x + 10$

Step 2) Substitute $- 4 x + 10$ for $y$ in the first equation and solve for $x$:

$3 x - 5 y = 53$ becomes:

$3 x - 5 \left(- 4 x + 10\right) = 53$

$3 x - \left(5 \cdot - 4 x\right) - \left(5 \cdot 10\right) = 53$

$3 x - \left(- 20 x\right) - 50 = 53$

$3 x + 20 x - 50 = 53$

$23 x - 50 = 53$

$23 x - 50 + \textcolor{red}{50} = 53 + \textcolor{red}{50}$

$23 x - 0 = 103$

$23 x = 103$

$\frac{23 x}{\textcolor{red}{23}} = \frac{103}{\textcolor{red}{23}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{23}}} x}{\cancel{\textcolor{red}{23}}} = \frac{103}{23}$

$x = \frac{103}{23}$

Step 3) Substitute $\frac{103}{23}$ for $x$ in the solution to the second equation at the end of Step 1 and calculate $y$

$y = - 4 x + 10$ becomes:

$y = \left(- 4 \cdot \frac{103}{23}\right) + 10$

$y = - \frac{412}{23} + \left(10 \cdot \frac{23}{23}\right)$

$y = - \frac{412}{23} + \frac{230}{23}$

$y = - \frac{182}{23}$

The solution is: $x = \frac{103}{23}$ and $y = - \frac{182}{23}$