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7.4: Solving Systems of Equations by Multiplication

Difficulty Level: 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.


Terms introduced in this lesson:

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*}7x+4y=175x2y=11samex27x+4y=1710x4y=22

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*}2x5y11x+3y=8=15

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

\begin{align*}2x-5y =8 && \xrightarrow{x11} && 22x-55y = 88 \\ -11x+3y =15 && \xrightarrow{x2} && -22x+6y = 30\end{align*}2x5y=811x+3y=15x11x222x55y=8822x+6y=30


\begin{align*}& \quad 22x-55y \ = 88 \\ & -22x+6y \ \ = 30 \\ & \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} \\ & \quad 0x-49y \ \ = 118\end{align*}22x55y =8822x+6y  =300x49y  =118

Divide by \begin{align*}-49\end{align*}49. Therefore, \begin{align*}y = -\frac{118}{49}\end{align*}y=11849.

To find \begin{align*}x\end{align*}x, eliminate \begin{align*}y\end{align*}y next.

\begin{align*}2x-5y =8 && \xrightarrow{x3} && 6x-15y = 24 \\ -11x+3y =15 && \xrightarrow{x5} && -55x+15y = 75\end{align*}2x5y=811x+3y=15x3x56x15y=2455x+15y=75


\begin{align*}& \quad 6x \ - \ 15y \ = 24\\ & -55x+15y \ = 75 \\ & \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} \\ & -49x + \ 0y \ \ = 99\end{align*}6x  15y =2455x+15y =7549x+ 0y  =99

Divide by \begin{align*}-49\end{align*}49. Therefore, \begin{align*}x= -\frac{99}{49}\end{align*}x=9949.

Answer: The solution to the system is \begin{align*}\left (-\frac{99}{49},-\frac{118}{49} \right )\end{align*}(9949,11849).

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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