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# 7.4: Solving Systems of Equations by Multiplication

Difficulty Level: At Grade Created by: CK-12

## Learning objectives

• 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.

## Introduction

We have now learned three techniques for solving systems of equations.

• Graphing, Substitution and Elimination (through addition and subtraction).

Each one of these methods has both strengths and weaknesses.

• Graphing is a good technique for seeing what the equations are doing, and when one service is less expensive than another, but graphing alone to find a solution can be imprecise and may not be good enough when an exact numerical solution is needed.
• Substitution is a good technique when one of the coefficients in an equation is +1 or -1, but can lead to more complicated formulas when there are no unity coefficients.
• Addition or Subtraction is ideal when the coefficients of either \begin{align*}x\end{align*} or \begin{align*}y\end{align*} match in both equations, but so far we have not been able to use it when coefficients do not match.

In this lesson, we will again look at the method of elimination that we learned in Lesson 7.3. However, the equations we will be working with will be more complicated and one can not simply add or subtract to eliminate one variable. Instead, we will first have to multiply equations to ensure that the coefficients of one of the variables are matched in the two equations.

Quick Review: Multiplying Equations

Consider the following questions

1. If 10 apples cost $5, how much would 30 apples cost? 2. If 3 bananas plus 2 carrots cost$4, how much would 6 bananas plus 4 carrots cost?

You can look at the first equation, and it should be obvious that each apple costs $0.50. 30 apples should cost$15.00.

7. A baker sells plain cakes for $7 or decorated cakes for$11. On a busy Saturday the baker started with 120 cakes, and sold all but three. His takings for the day were 991. How many plain cakes did he sell that day, and how many were decorated before they were sold? 8. Twice John’s age plus five times Claire’s age is 204. Nine times John’s age minus three times Claire’s age is also 204. How old are John and Claire? ## Review Answers 1. \begin{align*}x=0, y = -1.5\end{align*} 2. \begin{align*}x=3, y=5\end{align*} 3. \begin{align*}x=1.25, y=1.25\end{align*} 4. \begin{align*}x=\frac{2}{3}, y=\frac{3}{5}\end{align*} 5. \begin{align*}x=-\frac{8}{7}, y=-\frac{13}{7}\end{align*} 6. \begin{align*}x=-\frac{3}{46}, y=\frac{39}{46}\end{align*} 1. \begin{align*}x = -9, y= -3\end{align*} 2. \begin{align*}x=1, y=5\end{align*} 3. \begin{align*}x=4, y=3\end{align*} 4. \begin{align*}x=-2, y=3\end{align*} 5. \begin{align*}x=\frac{13}{8}, y=\frac{15}{32}\end{align*} 6. \begin{align*}x=-\frac{13}{25}, y=-\frac{17}{50}\end{align*} 1. \begin{align*}A = 114^{\circ}, B = 66^{\circ}\end{align*} 2. 30 liters of 5%, 70 liters of 15% 3. 51 yards and 99 yards 4.68, 750 in Company A, \$31, 250 in Company B
5. 74 plain, 43 decorated
6. John is 32, Claire is 28

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