Intermediate Algebra/Systems of Equations By Algebra

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[edit] 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.

[edit] Addition

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:
6x + 3y = 42
2x − 3y = 22
We would simply add up the values in the two equations, canceling out y in the process.
8x = 64 This is the result of the initial addition.
x = 8 Simplify.
Now, all we have to do is substitute 8 for each occurrence of x,and solve for y.
6(8) + 3y = 42 Substitute the value of x.
48 + 3y = 42 Simplify.
3y = − 6 Subtract 48 from each side.
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.
3x + 8y = 48
x − 4y = 22
We would simply multiply the second equation throughout by 2 and get:
2x − 8y = 44 Then add up:
x = 4 Substitute:
(8) − 4y = 22
− 4y = 18

[edit] 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 is on its one side.
y = 3x + 1
x + 2y = 16
Here you can simply substitute the first algebraic expression that y equals in to the second.
x + 2(3x + 1) = 16
Now simply slove the problem
x + 6x + 2 = 16
7x + 2 = 16
7x + 2 − 2 = 16 − 2

x = 2
Then plug it into the equation you substituted earlier.
y = 3(2) + 1
y = 6 + 1
y = 7
To check your work simply plug both x and y into one part of your system.
x + 2y = 16
(2) + 2(7) = 16
16 = 16 check.