Systems of Equations (Elimination)

Here you’ll learn the final method for solving a system of equations, the elimination method. Below are the instructions for how to use this method, followed by an example of solving a system of equations with this method.

Rule

The Elimination Method

1.: Choose a variable that will be eliminated.
2.: Multiply the equations $I$ and $I I$ by the numbers that would make the variable you chose have the same coefficient, but with opposite signs. This will allow you to cancel them.
3.: Write the new equations below the first two.
4.: Add equation $I I$ to equation $I$ , and write the answer below these equations. Now, you’re left with one equation, with one unknown. Solve it.
5.: Put the answer you find into equation $I$ , and solve for the final variable.
6.: Write your answer using coordinates:
ANSWER: $(x, y) = (a, b)$

Example 1

Solve the system of equations $\begin{array}{l} I & y + 2 x = 1 \\ I I & 2 y - x = 2 \end{array}$

1.: Let’s decide to get rid of $y$ .
2.: Since you have $2 y$ in equation $I I$ and $y$ in equation $I$ , you have to multiply equation $I$ by $- 2$ to eliminate $y$ . In this case, you don’t have to multiply equation $I I$ by anything: $\begin{array}{l} I y + 2 x = 1 & | \cdot (- 2) \\ I I \underset{̲}{2 y - x = 2} \end{array}$
3.: Write the two equations again after applying the changes: $\begin{array}{l} - 2 y - 4 x & = - 2 \\ \underset{̲}{2 y - x} & \underset{̲}{= 2} \end{array}$
4.: Add the two equations together such that $y$ is eliminated, and solve for $x$ : $\begin{array}{l} - 2 y + 2 y - 4 x - x & = - 2 + 2 \\ 0 y - 5 x & = 0 \\ x & = 0 \end{array}$
5.: Put the answer into either equation $I$ or $I I$ . Here, you can choose whichever you like. Let’s pick equation $I$ in this case: $\begin{array}{l} y + 2 x & = 1 \\ y + 2 \times (0) & = 1 \\ y + 0 & = 1 \\ y & = 1 \end{array}$
6.: Write the answer using coordinates:
ANSWER: $(x, y) = (0, 1)$

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Systems of Equations (Substitution)

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Activities 9