# How do you solve the following system of equations?: 4x – 5y = 14 , 7x + 3y=10?

##### 1 Answer
Feb 7, 2016

$y = - \frac{58}{47}$
$x = \frac{92}{47}$

#### Explanation:

There are a few ways to solve these two equations. One way is by graphing the two and seeing where they intersect. Another way is by substitution, where you isolate one variable in one of the equations, then plugging in that value in place of the variable.

My favorite way is by elimination. That is where you set up the equations just like how you'd do solve a normal subtraction/addition problem, such as this :
$\textcolor{w h i t e}{. .} 3$
$+$
$\textcolor{w h i t e}{. .} 4$
$- - -$
Then you set it up so that two variables have the same coefficient so they can be added or subtracted to equal zero. After that you're left with 1 variable, which can be easily solved for.

Once you have the value for that variable, you plug it into one of the original equations and solve for the other variable. After you have both variables, you check your answers by plugging them into the second equation.

This might sound complicated, but we'll put into practice and go through it step by step.

First, we set it up like an addition/subtraction problem:
$\textcolor{w h i t e}{\ldots .} 4 x - 5 y = 14$
$- +$
$\textcolor{w h i t e}{\ldots .} 7 x + 3 y = 10$

Now normally we'd just pick a variable to solve for, say $y$, and then either subtract or add to get rid of $x$. But, if we look, neither adding nor subtracting would get rid of the $x$s; they don't have the same coefficients. So, we need to multiply the equations to make the coefficients of $x$ subtract out. I think I'll do it this way:

$\textcolor{red}{\cdot 7} \textcolor{w h i t e}{\ldots} \cancel{4 x} 28 x - \cancel{5 y} 35 y = \cancel{14} 98$
$\textcolor{w h i t e}{\ldots} -$
$\textcolor{red}{\cdot 4} \textcolor{w h i t e}{\ldots} \cancel{7 x} 28 x + \cancel{3 y} 12 y = \cancel{10} 40$

Now if we subtract the $x$s we are left with
$- 47 y = 58$, which we can solve to be $\textcolor{red}{y = - \frac{58}{47}}$.

Let's solve for $x$ now.

First we plug $y$ into, let's say, $7 x + 3 y = 10$. That becomes $7 x + \left(3 \cdot - \frac{58}{47}\right) = 10$. We can simplify that to $7 x - \frac{174}{47} = 10$, which becomes $7 x = \frac{664}{47}$ or $\textcolor{red}{x = \frac{92}{47}}$

Let's double check our work by plugging in our values for $x$ and $y$ into the other equation, $4 x - 5 y = 14$.

That looks like this:
$\left(4 \cdot \frac{92}{47}\right) - \left(5 \cdot - \frac{58}{47}\right) = 14$, which becomes $\frac{368}{47} - - \frac{290}{47} = 14$, or $\frac{658}{47} = 14$, which is just $14 = 14$.

Congratulations! We were correct! $y = - \frac{58}{47}$ and $x = \frac{92}{47}$. If you want to check this another way, graph both equations and see the coordinate pairs for where they intersect; they should be (equivalent to) $\left(\frac{92}{47} , - \frac{58}{47}\right)$