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# 7.4: Solving Linear Systems by Multiplication

Difficulty Level: At Grade Created by: CK-12

This chapter has provided three methods to solve systems: graphing, substitution, and elimination through addition and subtraction. As stated in each lesson, these methods have strengths and weaknesses. Below is a summary.

Graphing

$\checkmark$ A good technique to visualize the equations and when both equations are in slope-intercept form.

• Solving a system by graphing is often imprecise and will not provide exact solutions.

Substitution

$\checkmark$ Works particularly well when one equation is in standard form and the second equation is in slope-intercept form.

$\checkmark$ Gives exact answers.

• Can be difficult to use substitution when both equations are in standard form.

$\checkmark$ Works well when both equations are in standard form and the coefficients of one variable are additive inverses.

$\checkmark$ Answers will be exact.

• Can be difficult to use if one equation is in standard form and the other is in slope-intercept form.
• Addition or subtraction does not work if the coefficients of one variable are not additive inverses.

Although elimination by only addition and subtraction does not work without additive inverses, you can use the Multiplication Property of Equality and the Distributive Property to create additive inverses.

Multiplication Property and Distributive Property:

If ax+by=c$ax+by=c$, then m(ax+by)=m(c)$m(ax+by)=m(c)$ and m(ax+by)=m(c)(am)x+(bm)y=mc$m(ax+by)=m(c)\rightarrow(am)x+(bm)y=mc$

While this definition box may seem complicated, it really states you can multiply the entire equation by a particular value and then use the Distributive Property to simplify. The value you are multiplying is called a scalar.

Example: Solve the system {7x+4y=125x2y=11$\begin{cases} 7x+4y=12\\ 5x-2y=11 \end{cases}$.

Solution: Neither variable has additive inverse coefficients. Therefore, simply adding or subtracting the two equations will not cancel either variable. However, there is a relationship between the coefficients of the y$y-$variable.

4 is the additive inverse of2×(2)
.

By multiplying the second equation by the scalar 2, you will create additive inverses of y$y$. You can then add the equations.

\begin{cases}
7x+4y=12\\
2(5x-2y=11) \end{cases} & \rightarrow \quad \begin{cases}
7x+4y=12\\
10x-4y=22 \end{cases}

\text{Add the two equations.} && 17x&=34\\
\text{Divide by} \ 17. && x&=2

To find the y$y-$value, use the Substitution Property in either equation.

7(2)+4y&=12\\
14+4y&=12\\
4y&=-2\\
y&=-\frac{1}{2}

The solution to this system is (2,12)$\left ( 2, -\frac{1}{2} \right )$.

Example: Andrew and Anne both use the I-Haul truck rental company to move their belongings from home to the dorm rooms on the University of Chicago campus. I-Haul has a charge per day and an additional charge per mile. Andrew travels from San Diego, California, a distance of 2,060 miles in five days. Anne travels 880 miles from Norfolk, Virginia, and it takes her three days. If Anne pays $840 and Andrew pays$1,845.00, what does I-Haul charge:

a) per day?

b) per mile traveled?

Solution: Begin by writing a system of linear equations: one to represent Anne and the second to represent Andrew. Let x=$x=$ amount charged per day and y=$y=$ amount charged per mile.

{3x+880y=8405x+2060y=1845

There are no relationships seen between the coefficients of the variables. Instead of multiplying one equation by a scalar, we must multiply both equations by the least common multiple.

The least common multiple is the smallest value that is divisible by two or more quantities without a remainder.

Suppose we wanted to eliminate the variable x$x$ because the numbers are smaller to work with. The coefficients of x$x$ must be additive inverses of the least common multiple.

LCM of 3 and 5=15

\begin{cases}
-5(3x+880y=840)\\
3(5x+2060y=1845) \end{cases} & \rightarrow \quad \begin{cases}
-15x-4400y=-4200\\
15x+6180y=5535 \end{cases}

&\text{Adding the two equations yields:} && 1780y =1335\\

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Feb 22, 2012

Dec 11, 2014