# 7.5: Linear Systems with Multiplication

**Basic**Created by: CK-12

**Practice**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*}

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

**Substitution**

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

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

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

**Elimination by Addition or Subtraction**

\begin{align*}\checkmark\end{align*}**and** the coefficients of one variable are additive inverses.

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

- 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*}

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*}

The company charges $0.75 per mile.

To find the amount charged per day, use the Substitution Property and either equation.

\begin{align*}5x+2060(0.75)&=1845\\ 5x+1545&=1845\\ 5x+1545-1545&=1845-1545\\ 5x&=300\\ x&=60\end{align*}

I-Haul charges $60.00 per day and $0.75 per mile.

### Vocabulary

**Elimination method:** The purpose of the **elimination method** to solve a system is to cancel, or eliminate, a variable by either adding or subtracting the two equations. Sometimes the equations must be multiplied by scalars first, in order to cancel out a variable.

**Least common multiple:** The ** least common multiple** is the smallest value that is divisible by two or more quantities

**without a remainder**.

### Guided Practice

*Solve the system* \begin{align*}\begin{cases}
4x+7y=6\\
6x+5y=20 \end{cases}\end{align*}.

**Solution:**

Neither \begin{align*}x\end{align*} nor \begin{align*}y\end{align*} have additive inverse coefficients, but the \begin{align*}x\end{align*}-variables do share a common factor of 2. Thus we can most easily eliminate \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 factor not shared in the coefficient of \begin{align*}x\end{align*} in the other equation. We need to multiply the first equation by 3 and the second equation by 2, making one of them negative as well:

\begin{align*}\begin{cases} 3(4x+7y=6)\\ -2(6x+5y=20) \end{cases} & \rightarrow \quad \begin{cases} 12x+21y=18\\ -12x-10y=-40\end{cases}\end{align*}

\begin{align*}\text{Add the two equations.} && 11y&=-22\\ \text{Divide by} \ 11. && y&=-2\end{align*}

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

\begin{align*}4x+7(2)&=6\\ 4x+14&=6\\ 4x&=-8\\ x&=-2\end{align*}

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

### Practice

Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Solving Linear Systems by Multiplication (12:00)

Find the least common multiple of the values given.

- 5 and 7
- –11 and 6
- 15 and 8
- 7 and 12
- 2 and 17
- –3 and 6
- 6 and \begin{align*}\frac{1}{3}\end{align*}
- 3 and 111
- 9 and 14
- 5 and –5

List the scalar needed to create the additive inverse.

- ____multiply by 6 to create additive inverse of 12.
- ____multiply by 5 to create additive inverse of 35.
- ____multiply by –10 to create additive inverse of 80.
- ____multiply by –7 to create additive inverse of 63.
- What could you multiply 11 by to create the additive inverse of 121?
- What scalar could you multiply 4 by to create the additive inverse of –16?

Solve the following systems using multiplication.

- \begin{align*}&5x-10y=15\\ &3x-2y=3\end{align*}
- \begin{align*}&5x-y=10\\ &3x-2y=-1\end{align*}
- \begin{align*}&5x+7y=15\\ &7x-3y=5\end{align*}
- \begin{align*}&9x+5y=9\\ &12x+8y=12.8\end{align*}
- \begin{align*}&4x-3y=1\\ &3x-4y=4\end{align*}
- \begin{align*}&7x-3y=-3\\ &6x+4y=3\end{align*}

In 23 – 28, solve the systems using any method.

- \begin{align*}&x=3y\\ &x-2y=-3\end{align*}
- \begin{align*}&y=3x+2\\ &y=-2x+7\end{align*}
- \begin{align*}&5x-5y=5\\ &5x+5y=35\end{align*}
- \begin{align*}&y=-3x-3\\ &3x-2y+12=0\end{align*}
- \begin{align*}&3x-4y=3\\ &4y+5x=10\end{align*}
- \begin{align*}&9x-2y=-4\\ &2x-6y=1\end{align*}
- Supplementary angles are two angles whose sum is \begin{align*}180^\circ\end{align*}. Angles \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are supplementary angles. The measure of angle \begin{align*}A\end{align*} is \begin{align*}18^\circ\end{align*} less than twice the measure of angle \begin{align*}B\end{align*}. Find the measure of each angle.
- A 150-yard pipe is cut to provide drainage for two fields. If the length of one piece is three yards less than twice the length of the second piece, what are the lengths of the two pieces?
- Mr. Stein invested a total of $100,00 in two companies for a year. Company A’s stock showed a 13% annual gain, while Company B showed a 3% loss for the year. Mr. Stein made an overall 8% return on his investment over the year. How much money did he invest in each company?
- 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?
- 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?

#### Quick Quiz

- Is (–3, –5) a solution to the system \begin{align*}\begin{cases} -3y=3x+6\\ y=-3x+4 \end{cases}\end{align*}?
- Solve the system: \begin{align*}\begin{cases} y=6x+17\\ y=7x+20 \end{cases}\end{align*}.
- Joann and Phyllis each improved their flower gardens by planting daisies and carnations. Joann bought 10 daisies and 4 carnations and paid $52.66. Phyllis bought 3 daisies and 6 carnations and paid $43.11 How much is each daisy? How much is each carnation?
- Terry’s Rental charges $49 per day and $0.15 per mile to rent a car. Hurry-It-Up charges a flat fee of $84 per day to rent a car. Write these two companies' charges in equation form and use the system to determine at what mileage the two companies will charge the same for a one-day rental.

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Least Common Multiple

The least common multiple of two numbers is the smallest number that is a multiple of both of the original numbers.elimination method

The purpose of the**elimination method**to solve a system is to cancel, or eliminate, a variable by either adding or subtracting the two equations. Sometimes the equations must be multiplied by scalars first, in order to cancel out a variable.

elimination

The elimination method for solving a system of two equations involves combining the two equations in order to produce one equation in one variable.Least Common Denominator

The least common denominator or lowest common denominator of two fractions is the smallest number that is a multiple of both of the original denominators.### Image Attributions

Here you'll learn how to multiply by scalars in order to use elimination to solve systems of linear equations.