# How do I use substitution to find the solution of the system of equations 4x+3y=7 and 3x+5y=8?

Sep 17, 2014

To use substitution, you would solve one of the equations for x (or y), and then plug it into the other equation for x(or y). You'll find that you will end up with one equation and one variable, which you can solve quite simply using algebra. When you get that answer, you can plug it back into the original equation you started with to get the other variable.

Since these are simultaneous equations, you can actually solve for x or y in either equation- You'll get the same answer either way.

Solution:
Equation 1: $4 x + 3 y = 7$
Equation 2: $3 x + 5 y = 8$

Solving Equation 1 for $x$, we get:
$x = \frac{7 - 3 y}{4}$

Plugging $x$ into Equation 2 :
$3 \left(\frac{7 - 3 y}{4}\right) + 5 y = 8$

Now we only have one variable that we can solve for. Solving Equation 2 :

$\left(\frac{21 - 9 y}{4}\right) + \frac{20 y}{4} = 8$ (I got the $\frac{20 y}{4}$ by multiplying the numerator of 5 by 4 to find a common denominator so I can add it to the other fraction)
$21 - 9 y + 20 y = 32$
$11 y = 11$
$y = 1$

We've solved for one of the variables! To solve for the other, we plug this back into Equation 1 to solve for $x$. So, since $y = 1$

$4 x + 3 \left(1\right) = 7$
$4 x = 4$
$x = 1$

And we've solved for the other variable. Interestingly, both the x and y values are 1, so $x = y = 1$. You can always check your answer by plugging these values back into the original equations and seeing if they satisfy them, which they do:

$4 \left(1\right) + 3 \left(1\right) = 7$
and
$3 \left(1\right) + 5 \left(1\right) = 8$

Remember that you don't have to follow this exact method to solve this. You can solve either equation for x or y and plug into the other and then plug back into the one you started with, and you should get the same answer.