# Systems of Equations (Elimination)

All linear equations graph as lines. A system of equations is more than one equation, so more than one line. The \"solution\" to that system of equations is where the lines cross.

Most lines intersect (or cross) once, so they have **one solution**, which is written as a coordinate. The solution of the system of lines graphed below is approximately (-3, -1)

If the two equations actually represent the same line (once you put the equations into $y=mx+b$ form, you should be able to see that they are identical), the system has** infinite solutions**. In the system below, the black line and the blue line are the same line, and both go on forever, so there are infinite solutions.

If the two equations represent parallel lines (you will see that they have the same slope), the system has **no solutions**. The lines below will never cross. The system below has no solutions.

As you can see, you can solve a system of equations by graphing. The point where the lines intersect is the answer.

But, graphs can be imperfect and lines do not always intersect somewhere where it is easy to read the coordinate, so there are two ways to solve systems of equations algebraically.

One way is through substitution (see lesson Algebra: Solve and Substitute).

Another, often cleaner, way is through elimination.

To use elimination, you arrange your two equations in the same way (variables in the same order, on the same side of the equal sign). Then you'll add those two equations together, with the goal of \"eliminating\" a variable, so you can solve for the remaining variable.

*Example:*

$$\require{cancel}\eqalign{&y=6+8x\quad \text{First equation}\\&2x - y=12\quad \text{Second equation}\\&\text{You will need to use algebra rules to get these questions in the same format, with variables lined up}}$$

$$\eqalign{&\left. \begin{array}{rcl} &&y&&&=6+8x\\ &&-8x &&& \qquad -8x\\&&-8x+y&&&=6\\&&2x+y&&&=12& \\\hline \end{array}\right\}\quad \text{Rewrite first equation to line up like terms} \\&\left. \begin{array}{rcl} &&-8x\cancel{+y}&&&=6\\ +&&2x\cancel{-y}&&&=12 \\\hline &&-6x&&&=18 \end{array}\right\} \quad \text{Add like terms together working vertically}\\&\left. \begin{array}{rcl} &&-6x&&&=18\\ &&x&&&=-3 \end{array}\right\} \quad \text{Solve for x}\\&\left. \begin{array}{rcl} &&2x-y&&&=12\\ &&2(-3)-y&&&=12 \\\hline &&-6-y&&&=12\\&&-y&&&=18\\&&y&&&=-18 \end{array}\right\} \quad \text{Plug x into one equation to solve for y}\\\text{The solution to this system of equations is (-3, -18)}}$$

The two equations in the example above have **one solution**. They intersect at (2,0). That problem shows you the basics of elimination, but sometimes you have to work on the equations for a while to get a variable to \"eliminate\" itself nicely.

*Example:*

$$\eqalign{&y=6+8x\quad \text{First equation}\\&2x - y=12\quad \text{Second equation}\\&\text{You will need to use algebra rules to get these questions in the same format, with variables lined up}}$$

$$\eqalign{&\left. \begin{array}{rcl} &&y&&&=6+8x\\ &&-8x &&& \qquad -8x\\&&-8x+y&&&=6\\&&2x+y&&&=12& \\\hline \end{array}\right\}\quad \text{Rewrite first equation to line up like terms} \\&\left. \begin{array}{rcl} &&-8x\cancel{+y}&&&=6\\ +&&2x\cancel{-y}&&&=12 \\\hline &&-6x&&&=18 \end{array}\right\} \quad \text{Add like terms together working vertically}\\&\left. \begin{array}{rcl} &&-6x&&&=18\\ &&x&&&=-3 \end{array}\right\} \quad \text{Solve for x}\\&\left. \begin{array}{rcl} &&2x-y&&&=12\\ &&2(-3)-y&&&=12 \\\hline &&-6-y&&&=12\\&&-y&&&=18\\&&y&&&=-18 \end{array}\right\} \quad \text{Plug x into one equation to solve for y}\\\text{The solution to this system of equations is (-3, -18)}}$$

The two equations in the example above have **one solution**. They intersect at (-3,-18).

Finally, sometimes even once equations are organized in the same way, you have to multiply one or both of the equations to get a variable to "eliminate" itself when you add or combine the equations.

*Example:*

$$\eqalign{&y=6+8x\quad \text{First equation}\\&2x - y=12\quad \text{Second equation}\\&\text{You will need to use algebra rules to get these questions in the same format, with variables lined up}}$$

$$\eqalign{&\left. \begin{array}{rcl} &&y&&&=6+8x\\ &&-8x &&& \qquad -8x\\&&-8x+y&&&=6\\&&2x+y&&&=12& \\\hline \end{array}\right\}\quad \text{Rewrite first equation to line up like terms} \\&\left. \begin{array}{rcl} &&-8x\cancel{+y}&&&=6\\ +&&2x\cancel{-y}&&&=12 \\\hline &&-6x&&&=18 \end{array}\right\} \quad \text{Add like terms together working vertically}\\&\left. \begin{array}{rcl} &&-6x&&&=18\\ &&x&&&=-3 \end{array}\right\} \quad \text{Solve for x}\\&\left. \begin{array}{rcl} &&2x-y&&&=12\\ &&2(-3)-y&&&=12 \\\hline &&-6-y&&&=12\\&&-y&&&=18\\&&y&&&=-18 \end{array}\right\} \quad \text{Plug x into one equation to solve for y}\\\text{The solution to this system of equations is (-3, -18)}}$$

The two equations in the example above have **one solution**. They intersect at (-3,-18).

Finally, sometimes even once equations are organized in the same way, you have to multiply one or both of the equations to get a variable to \"eliminate\" itself when you add or combine the equations.

*Example:*

$$\eqalign{&9x-2y=14\quad \text{First equation}\\&3x - y=6\quad \text{Second equation}\\&\text{You'll need to multiple an entire equation by something to get a variable to cancel}}$$

$$\eqalign{&\left. \begin{array}{rcl} &&9x-2y&&&=14\\ (-3)(&&3x-y&&&=6)\\&&-9x+3y&&&=-18& \\\hline \end{array}\right\}\quad \text{Multiply second equation by -3} \\&\left. \begin{array}{rcl} &&\cancel{9x}-2y&&&=14\\ +&&\cancel{-9x}+3y&&&=-18 \\\hline &&y&&&=-4 \end{array}\right\} \quad \text{Add like terms together working vertically}\\&\left. \begin{array}{rcl} &&3x-y&&&=6\\ &&3x-(-4)&&&=6 \\\hline &&3x&&&=10\\&&x&&&=\dfrac{10}{3} \end{array}\right\} \quad \text{Plug x into one equation to solve for y}\\\text{The solution to this system of equations is} (\dfrac{10}{3}, -4)}$$

The two equations in the example above have** one solution**. They intersect at (10/3,-4).

Remember, before you go through all of the steps above, check two things:

- In simplest form, are the equations the same? If so, they are the same line:
**infinite solutions.** - In simplest form, are the equations different but have the same slope? If so, they are parallel:
**no solution.**

#### Practice Problems:

## Algebra: Systems of Equations (Elimination)

$$\eqalign{2x+3y&=6\\4x-3y&=3}$$

1) What is the x value of the intersection of the equations above?

2) What is the y value of the intersection of the equations above?$$\eqalign{x+2y&=8\\-x-3y&=4}$$

3) What is the x value of the intersection of the equations above?

4) What is the y value of the intersection of the equations above?$$\eqalign{5x-4y&=2\\-x-y&=2}$$

5) What is the x value of the intersection of the equations above?

6) What is the y value of the intersection of the equations above?$$\eqalign{y&=2+3x\\-2y&=4x-2}$$

7) What is the solution of the system of equations above?

$$\eqalign{3x-y&=-2\\-x&=6y-12}$$

8) What is the solution of the system of equations above?

$$\eqalign{4x-4y&=4\\y&=x+2}$$

9) What is the solution of the system of equations above?

$$\eqalign{-2y&=1-3x\\-2x&=3y+2}$$

10) What is the solution of the system of equations above?

#### Answer Key: