# Intermediate Algebra/Systems of Equations By Algebra

## Solving Systems of Linear Equations by Using Algebra

Generally, you're not going to want to solve a system using graphs, simply because it takes too much time. There are two algebraic methods for solving systems of linear equations.

The ideal situation for the Addition method (also known as Elimination method) is one in which a variable in the two equations has opposite coefficients. For instance:
${\displaystyle 6x+3y=42}$
${\displaystyle 2x-3y=22}$
We would simply add up the values in the two equations, canceling out ${\displaystyle y}$ in the process.
${\displaystyle 8x=64}$ This is the result of the initial addition.
${\displaystyle x=8}$ Simplify.
Now, all we have to do is substitute ${\displaystyle 8}$ for each occurrence of ${\displaystyle x}$,and solve for ${\displaystyle y}$.
${\displaystyle 6(8)+3y=42}$ Substitute the value of ${\displaystyle x}$.
${\displaystyle 48+3y=42}$ Simplify.
${\displaystyle 3y=-6}$ Subtract 48 from each side.
${\displaystyle y=-2}$ Divide each side by 3.

However, even if the variables don't easily cancel out, simply just try with constant multiplications and so on.
${\displaystyle 3x+8y=48}$
${\displaystyle x-4y=22}$
We would simply multiply the second equation throughout by 2 and get:
${\displaystyle 2x-8y=44}$ Then add up:
${\displaystyle x=4}$ Substitute:
${\displaystyle (8)-4y=22}$
${\displaystyle -4y=18}$
${\displaystyle y=-{\frac {9}{2}}}$

In some occasions, you may need to multiply both sides. For example:

${\displaystyle 3y+2x=5}$
${\displaystyle 4y+3x=10}$

In this case, we will multiply the first equation by three and the second equation by two.

${\displaystyle 9y+6x=15}$
${\displaystyle 8y+6x=20}$

${\displaystyle y=-5}$

${\displaystyle 9\times -5+6x=20}$
${\displaystyle -45+6x=20}$
${\displaystyle 6x=85}$
${\displaystyle x=14{\frac {1}{6}}}$

### Substitution

This is another method to solve a system of linear equations. This is ideal if one of the equations is laid out where one variable has a coefficient of one or negative one.
${\displaystyle y=3x+1}$
${\displaystyle x+2y=16}$
Here you can simply substitute the first algebraic expression that y equals in to the second.
${\displaystyle x+2(3x+1)=16}$
Now simply slove the problem
${\displaystyle x+6x+2=16}$
${\displaystyle 7x+2=16}$
${\displaystyle 7x+2-2=16-2}$
${\displaystyle {\frac {7x}{7}}={\frac {14}{7}}}$
${\displaystyle x=2}$
Then plug it into the equation you substituted earlier.
${\displaystyle y=3(2)+1}$
${\displaystyle y=6+1}$
${\displaystyle y=7}$
To check your work simply plug both x and y into one part of your system.
${\displaystyle x+2y=16}$
${\displaystyle (2)+2(7)=16}$
${\displaystyle 16=16}$ check.

Example where variable is not on one side:

${\displaystyle x+y=9}$
${\displaystyle 3x+5y=25}$

Switch first equation so x is on one side

${\displaystyle x=9-y}$

Substitute

${\displaystyle 3(9-y)+5y=25}$

Distribute and solve

${\displaystyle 27-3y+5y=25}$
${\displaystyle 2y=-2}$

${\displaystyle y=-1}$
${\displaystyle x=10}$