Linear Systems with Multiplication

Multiply as needed to set coefficients opposite, add to cancel variables

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Linear Systems with Multiplication

Suppose that you want to buy some guppies and rainbowfish for your aquarium. If you buy 10 guppies and 15 rainbowfish, it will cost you $90, and if you buy 15 guppies and 10 rainbowfish, it will cost you$85. How much do guppies and rainbowfish each cost? Can you set up a system of equations to find out? How would you go about solving it? In this Concept, you'll learn about solving systems of linear equations by multiplication so that you can find the solution to this type of problem.

Watch This

Multimedia Link: For even more practice, we have this video. One common type of problem involving systems of equations (especially on standardized tests) is “age problems.” In the following video, the narrator shows two examples of age problems, one involving a single person and one involving two people. Khan Academy Age Problems (7:13)

For more help with solving systems by eliminating a variable, visit this site: http://www.brightstorm.com/math/algebra/solving-systems-of-equations/solving-systems-of-equations-using-elimination - Brightstorm video. You may need to set up a free account with Brightstorm to finish watching the video.

Guidance

The previous Concepts have provided three methods to solve systems: graphing, substitution, and elimination through addition and subtraction. As stated in each Concept, these methods have strengths and weaknesses. Below is a summary.

Graphing

\begin{align*}\checkmark\end{align*} 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

\begin{align*}\checkmark\end{align*} Works particularly well when one equation is in standard form and the second equation is in slope-intercept form.

\begin{align*}\checkmark\end{align*} Gives exact answers.

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

\begin{align*}\checkmark\end{align*} Works well when both equations are in standard form and the coefficients of one variable are additive inverses.

\begin{align*}\checkmark\end{align*} 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 \begin{align*}ax+by=c\end{align*}, then \begin{align*}m(ax+by)=m(c)\end{align*} and \begin{align*}m(ax+by)=m(c)\rightarrow(am)x+(bm)y=mc\end{align*}

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

Example A

Solve the system \begin{align*}\begin{cases} 7x+4y=12\\ 5x-2y=11 \end{cases}\end{align*}.

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 \begin{align*}y-\end{align*}variable.

\begin{align*}4 \ is \ the \ additive \ inverse \ of-2 \times (2)\end{align*}.

By multiplying the second equation by the scalar 2, you will create additive inverses of \begin{align*}y\end{align*}. You can then add the equations.

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

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

To find the \begin{align*}y-\end{align*}value, use the Substitution Property in either equation.

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

The solution to this system is \begin{align*}\left ( 2, -\frac{1}{2} \right )\end{align*}.

Example B

Solve the system \begin{align*}\begin{cases} 3x+4y=-25\\ 2x-3y=6 \end{cases}\end{align*}.

Solution:

Not only does neither variable have additive inverse coefficients, but there also isn't a relationship between the coefficients of either variable. In other words, the coefficients of each variable do not share any common factors. Therefore, we can try to eliminate either variable first. We will show how to solve this problem by eliminating \begin{align*}x\end{align*}.

In order to make \begin{align*}x\end{align*} have the same coefficient in each equation, we must multiply one equation by the coefficient of \begin{align*}x\end{align*} in the other equation. We need to multiply the first equation by 2 and the second equation by 3. If we make one of those numbers negative, we can easily eliminate \begin{align*}x\end{align*}. It does not matter which one we make negative:

\begin{align*}\begin{cases} 2(3x+4y=-25)\\ -3(2x-3y=6) \end{cases} & \rightarrow \quad \begin{cases} 6x+8y=-50\\ -6x+9y=-18 \end{cases}\end{align*}

\begin{align*}\text{Add the two equations.} && 17y&=68\\ \text{Divide by} \ 17. && y&=-4\end{align*}

To find the \begin{align*}x-\end{align*}value, use the Substitution Property in either equation.

\begin{align*}2x-3(-4)&=6\\ 2x+12&=6\\ 2x&=-6\\ x&=-3\end{align*}

The solution to this system is \begin{align*} (-3,-4)\end{align*}.

Example C

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 \begin{align*}x=\end{align*} amount charged per day and \begin{align*}y=\end{align*} amount charged per mile.

\begin{align*}\begin{cases} 3x+880y=840\\ 5x+2060y=1845 \end{cases}\end{align*}

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 \begin{align*}x\end{align*} because the numbers are smaller to work with. The coefficients of \begin{align*}x\end{align*} must be additive inverses of the least common multiple.

\begin{align*}LCM \ of \ 3 \ and \ 5=15\end{align*}

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

\begin{align*}&\text{Adding the two equations yields:} && 1780y =1335\\ &\text{Divide by} 1780: && \qquad \quad \ y=0.75\end{align*}

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