View source for 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.



=== 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:<br>
<math>6x + 3y = 42</math><br>
<math>2x - 3y = 22</math><br>
We would simply add up the values in the two equations, canceling out <math>y</math> in the process.<br>
<math>8x = 64</math> This is the result of the initial addition.<br>
<math>x = 8</math> Simplify.<br>
Now, all we have to do is substitute <math>8</math> for each occurrence of <math>x</math>,and solve for <math>y</math>.<br>
<math>6(8) + 3y = 42</math> Substitute the value of <math>x</math>.<br>
<math>48 + 3y = 42</math> Simplify.<br>
<math>3y = -6</math> Subtract 48 from each side.<br>
<math>y = -2</math> Divide each side by 3.<br>

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

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

<math>3y + 2x = 5</math><br>
<math>4y + 3x = 10</math>

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

<math>9y + 6x = 15</math><br>
<math>8y + 6x = 20</math>

<math>y=-5</math>

<math>9 \times -5 + 6x = 20</math><br>
<math>-45 +6x = 20</math><br>
<math>6x=85</math><br>
<math>x = 14 \frac {1} {6}</math>

=== 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.<br/>
<math>y = 3x + 1</math><br/>
<math>x + 2y = 16</math><br/>
Here you can simply substitute the first algebraic expression that y equals in to the second.<br/>
<math>x + 2(3x + 1) = 16</math><br/>
Now simply slove the problem<br/>
<math>x + 6x + 2 = 16</math><br/>
<math>7x + 2 = 16</math><br/>
<math>7x + 2 - 2= 16 -2</math><br/>
<math>\frac{7x}{7}  = \frac{14}{7}</math><br/>
<math>x = 2</math><br/>
Then plug it into the equation you substituted earlier.<br/>
<math>y = 3(2) + 1</math><br/>
<math>y = 6 + 1</math><br/>
<math>y = 7</math><br/>
To check your work simply plug both x and y into one part of your system.<br/>
<math>x + 2y = 16</math><br/>
<math>(2) + 2(7) = 16</math><br/>
<math>16 = 16</math> check.<br/>



Example where variable is not on one side:

<math>x + y = 9</math><br>
<math>3x + 5y = 25</math>


Switch first equation so x is on one side

<math>x = 9 - y</math>


Substitute

<math>3(9-y) + 5y = 25</math>


Distribute and solve

<math>27 - 3y + 5y = 25</math><br>
<math>2y = -2</math>

<math>y = -1</math><br>
<math>x = 10</math>

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