# 7.4: Solving Systems of Equations by Multiplication

**At Grade**Created by: CK-12

## Learning Objectives

At the end of this lesson, students will be able to:

- Solve a linear system by multiplying one equation.
- Solve a linear system of equations by multiplying both equations.
- Compare methods for solving linear systems.
- Solve real-world problems using linear systems by any method.

## Vocabulary

Terms introduced in this lesson:

- scalar
- lowest common multiple

## Teaching Strategies and Tips

Use the introduction to convince students that the elimination method applies to *any* linear system because one or both equations can be multiplied by a constant resulting in a “new” pair of equations with matching coefficients.

In Example 1, some students have trouble keeping track of the multipliers for each equation. Try using a visual:

\begin{align*}7x + 4y = 17 && \text{same} && 7x + 4y = 17 \\
5x - 2y = 11 && \xrightarrow{x2} && 10x - 4y = 22\end{align*}

In Example 2,

- Point out that the distance covered is the same in both directions, so the \begin{align*}400 \;\mathrm{yards}\end{align*}
400yards is unnecessary information. - Remind students to back-substitute to complete the problem.

Use Example 3 to demonstrate a system with no matching coefficients and no coefficients that are multiples of others.

- Remind students how to find the LCM of two numbers.
- Suggest that students find the LCM of the “smaller pair” of coefficients, \begin{align*}3\end{align*}
3 and \begin{align*}5\end{align*}5 , instead of \begin{align*}880\end{align*}880 and 1845.

Teachers may decide to forgo back-substitution and instead teach elimination of the second variable (“double elimination”).

Additional Example:

*Solve the system using multiplication.*

\begin{align*}2x-5y & =8 \\
-11x+3y & =15\end{align*}

Solution by “double elimination”. Eliminate \begin{align*}x\end{align*}

\begin{align*}2x-5y =8 && \xrightarrow{x11} && 22x-55y = 88 \\
-11x+3y =15 && \xrightarrow{x2} && -22x+6y = 30\end{align*}

Add.

\begin{align*}& \quad 22x-55y \ = 88 \\
& -22x+6y \ \ = 30 \\
& \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} \\
& \quad 0x-49y \ \ = 118\end{align*}

Divide by \begin{align*}-49\end{align*}

To find \begin{align*}x\end{align*}

\begin{align*}2x-5y =8 && \xrightarrow{x3} && 6x-15y = 24 \\
-11x+3y =15 && \xrightarrow{x5} && -55x+15y = 75\end{align*}

Add.

\begin{align*}& \quad 6x \ - \ 15y \ = 24\\ & -55x+15y \ = 75 \\ & \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} \\ & -49x + \ 0y \ \ = 99\end{align*}

Divide by \begin{align*}-49\end{align*}. Therefore, \begin{align*}x= -\frac{99}{49}\end{align*}.

Answer: The solution to the system is \begin{align*}\left (-\frac{99}{49},-\frac{118}{49} \right )\end{align*}.

The advantage of “double elimination” is that the fraction does not need to be back-substituted.

## Error Troubleshooting

General Tip: Students forget to multiply *every* term in an equation by the scalar.

General Tip: Encourage students to eliminate the variable whose coefficients in both equations have the smallest LCM.

Remind students in *Review Problem* 2e to align the variables column-wise.

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