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# 7.4: Solving Systems of Equations by Multiplication

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:

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

In Example 2,

• Point out that the distance covered is the same in both directions, so the $400 \;\mathrm{yards}$ 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, $3$ and $5$, instead of $880$ and 1845.

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

Solve the system using multiplication.

$2x-5y & =8 \\-11x+3y & =15$

Solution by “double elimination”. Eliminate $x$ first.

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

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

Divide by $-49$. Therefore, $y = -\frac{118}{49}$.

To find $x$, eliminate $y$ next.

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

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

Divide by $-49$. Therefore, $x= -\frac{99}{49}$.

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

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.

Feb 22, 2012

Aug 22, 2014