The cost of two cell phone plans can be written as a system of equations based on the number of minutes used and the base monthly rate. As a consumer, it would be useful to know when the two plans cost the same and when is one plan cheaper.

Plan A costs $40 per month plus $0.10 for each minute of talk time.

Plan B costs $25 per month plus $0.50 for each minute of talk time.

Plan B has a lower starting cost, but since it costs more per minute, it may not be the right plan for someone who likes to spend a lot of time on the phone. When do the two plans cost the same amount?

#### Watch This

http://www.youtube.com/watch?v=ova8GSmPV4o James Sousa: Solving Systems of Equations by Elimination

#### Guidance

There are many ways to solve a system that you have learned in the past including substitution and graphical intersection. Here you will focus on solving using elimination because the knowledge and skills used will transfer directly into using matrices.

When solving a system, the first thing to do is to count the number of variables that are missing and the number of equations. The number of variables needs to be the same or fewer than the number of equations. Two equations and two variables can be solved, but one equation with two variables cannot.

Get into the habit of always writing systems in standard form:

**Example A**

Solve the following system of equations:

**Solution: ** Here is a system of two equations and two variables in standard form. Notice that there is an

Equation 1:

Equation 2:

Strategically choose to eliminate

Add the two equations:

The value for

Equation 1:

Equation 2:

Adding the two equations:

The point

**Example B**

Solve the following system of equations:

**Solution: ** Scaling the first equation by -2 will allow the

The sum is:

You can substitute

The point

**Example C**

Solve the following system of equations:

**Solution: ** The strategy of elimination still applies. You can eliminate the

Add the equations together and solve for

Substitute into the second equation and solve for

The point

**Concept Problem Revisited**

Plan A costs $40 per month plus $0.10 for each minute of talk time.

Plan B costs $25 per month plus $0.50 for each minute of talk time.

If you want to find out when the two plans cost the same, you can represent each plan with an equation and solve the system of equations. Let

First you put these equations in standard form.

Then you scale the second equation by -1 and add the equations together and solve for

To solve for

The equivalent costs of plan A and plan B will occur at 37.5 minutes of talk time with a cost of $43.75.

#### Vocabulary

A ** system of equations** is two or more equations.

** Standard form** for the equation of a line is

To ** scale an equation** means to multiply every term by a constant.

#### Guided Practice

1. Solve the following system using elimination:

2. Solve the following system using elimination:

3. Solve the following system using elimination:

**Answers:**

1. Start by scaling both of the equations by

\begin{align*}70x+21y &= -112\\ 27x-21y &= 15\end{align*}

Add, solve for \begin{align*}x=-1\end{align*}, substitute and solve for \begin{align*}y\end{align*}.

Final Answer: (-1, -2)

2. Start by scaling the first equation by 7 and notice that the \begin{align*}y\end{align*} coefficient will immediately be eliminated when the equations are summed.

\begin{align*}35x-7y &= 154\\ -2x+7y &= 19\end{align*}

Add, solve for \begin{align*}x=\frac{173}{33}\end{align*}. Instead of substituting, practice eliminating \begin{align*}x\end{align*} by scaling the first equation by 2 and the second equation by 5.

\begin{align*}10x-2y &= 44\\ -10x+35y &= 95\end{align*}

Add, solve for \begin{align*}y\end{align*}.

Final Answer: \begin{align*}\left(\frac{173}{33}, \frac{139}{33}\right)\end{align*}

3. To eliminate \begin{align*}\frac{1}{y}\end{align*}, scale the first equation by 2 and the second equation by 5.

To eliminate \begin{align*}\frac{1}{x}\end{align*}, scale the first equation by -9 and the second equation by 11.

Final Answer: \begin{align*}\left(-\frac{1}{3}, 1 \right)\end{align*}

#### Practice

Solve each system of equations using the elimination method.

1. \begin{align*}x+y=-4; -x+2y=13\end{align*}

2. \begin{align*}\frac{3}{2}x-\frac{1}{2}y=\frac{1}{2}; -4x+2y=4\end{align*}

3. \begin{align*}6x+15y=1; 2x-y=19\end{align*}

4. \begin{align*}x-\frac{2y}{3}=\frac{-2}{3}; 5x-2y=10\end{align*}

5. \begin{align*}-9x-24y=-243; \frac{1}{2}x+y=\frac{21}{2}\end{align*}

6. \begin{align*}5x+\frac{28}{3}y=\frac{176}{3}; y+x=10\end{align*}

7. \begin{align*}2x-3y=50; 7x+8y=-10\end{align*}

8. \begin{align*}2x+3y=1; 2y=-3x+14\end{align*}

9. \begin{align*}2x+\frac{3}{5}y=3;\frac{3}{2}x-y=-5\end{align*}

10. \begin{align*}5x=9-2y;3y=2x-3\end{align*}

11. How do you know if a system of equations has no solution?

12. If a system of equations has no solution, what does this imply about the relationship of the curves on the graph?

13. Give an example of a system of two equations with two unknowns with an infinite number of solutions. Explain how you know the system has an infinite number of solutions.

14. Solve:

\begin{align*}12 \cdot \frac{1}{x}-18 \cdot \frac{1}{y} &= 4\\ 8 \cdot \frac{1}{x}+9 \cdot \frac{1}{y} &= 5\end{align*}

15. Solve:

\begin{align*}14 \cdot \frac{1}{x}-5 \cdot \frac{1}{y} &= -3\\ 7 \cdot \frac{1}{x}+2 \cdot \frac{1}{y} &= 3\end{align*}