# 7.5: Linear Systems with Multiplication

**At Grade**Created by: CK-12

**Practice**Linear Systems with Multiplication

### Linear Systems with Multiplication

So far, we’ve seen that the elimination method works well when the coefficient of one variable happens to be the same (or opposite) in the two equations. But what if the two equations don’t have any coefficients the same?

It turns out that we can still use the elimination method; we just have to *make* one of the coefficients match. We can accomplish this by multiplying one or both of the equations by a constant.

Here’s a quick review of how to do that. Consider the following questions:

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

If you look at the first equation, it should be obvious that each apple costs $0.50. So 30 apples should cost $15.00.

The second equation is trickier; it isn’t obvious what the individual price for either bananas or carrots is. Yet we know that the answer to question 2 is $8.00. How?

If we look again at question 1, we see that we can write an equation: \begin{align*}10a = 5\end{align*} (\begin{align*}a\end{align*} being the cost of 1 apple). So to find the cost of 30 apples, we *could* solve for \begin{align*}a\end{align*} and then multiply by 30—but we could also just multiply both sides of the equation by 3. We would get \begin{align*}30a = 15\end{align*}, and that tells us that 30 apples cost $15.

And we can do the same thing with the second question. The equation for this situation is \begin{align*}3b + 2c = 4\end{align*}, and we can see that we need to solve for \begin{align*}(6b + 4c)\end{align*}, which is simply 2 times \begin{align*}(3b + 2c)\end{align*}! So algebraically, we are simply multiplying the entire equation by 2:

\begin{align*}2(3b + 2c) &= 2 \cdot 4 && distribute \ and \ multiply:\\ 6b + 4c&= 8\end{align*}

So when we multiply an equation, all we are doing is multiplying every term in the equation by a fixed amount.

**Solving a Linear System by Multiplying One Equation**

If we can multiply every term in an equation by a fixed number (a **scalar**), that means we can use the addition method on a whole new set of linear systems. We can manipulate the equations in a system to ensure that the coefficients of one of the variables match.

This is easiest to do when the coefficient as a variable in one equation is a multiple of the coefficient in the other equation.

Solve the system:

\begin{align*}7x + 4y & = 17\\ 5x - 2y & = 11\end{align*}

You can easily see that if we multiply the second equation by 2, the coefficients of \begin{align*}y\end{align*} will be +4 and -4, allowing us to solve the system by addition:

*2 times equation 2:*

\begin{align*}& \qquad \qquad \qquad \ \ \quad 10x - 4y = 22 && now \ add \ to \ equation \ one:\\ & \qquad \qquad \quad \underline{\;\;\;+ \ \ (7x + 4y) = 17\;\;\;\;\;}\\ & \qquad \qquad \qquad \qquad \ \ \ \quad 17x = 34\\ \\ & divide \ by \ 17 \ to \ get: \qquad \ x = 2\end{align*}

Now simply substitute this value for \begin{align*}x\end{align*} back into equation 1:

\begin{align*}7 \cdot 2 + 4y &= 17 && since \ 7 \times 2 = 14, \ subtract \ 14 \ from \ both \ sides:\\ 4y &= 3 && divide \ by \ 4:\\ y &= 0.75\end{align*}

#### Converting a Word Problem into a System of Equations

Anne is rowing her boat along a river. Rowing downstream, it takes her 2 minutes to cover 400 yards. Rowing upstream, it takes her 8 minutes to travel the same 400 yards. If she was rowing equally hard in both directions, calculate, in yards per minute, the speed of the river and the speed Anne would travel in calm water.

Step one: first we convert our problem into equations. We know that *distance traveled* is equal to *speed* \begin{align*}\times\end{align*} *time*. We have two unknowns, so we’ll call the speed of the river \begin{align*}x\end{align*}, and the speed that Anne rows at \begin{align*}y\end{align*}. When traveling downstream, her total speed is her rowing speed plus the speed of the river, or \begin{align*}(x + y)\end{align*}. Going upstream, her speed is hindered by the speed of the river, so her speed upstream is \begin{align*}(x - y)\end{align*}.

Downstream Equation: \begin{align*}2(x + y) = 400\end{align*}

Upstream Equation: \begin{align*}8(x - y) = 400\end{align*}

Distributing gives us the following system:

\begin{align*}2x + 2y &= 400\\ 8x - 8y &= 400\end{align*}

Right now, we can’t use the method of elimination because none of the coefficients match. But if we multiplied the top equation by 4, the coefficients of \begin{align*}y\end{align*} would be +8 and -8. Let’s do that:

\begin{align*}& \quad \qquad \ 8x + 8y = 1,600\\ & \ \underline{\;\;\; + \ \ (8x - 8y) = 400\;\;\;}\\ & \quad \qquad \qquad 16x = 2,000\end{align*}

Now we divide by 16 to obtain \begin{align*}x = 125\end{align*}.

Substitute this value back into the first equation:

\begin{align*}2(125 + y) &= 400 && divide \ both \ sides \ by \ 2:\\ 125 + y &= 200 && subtract \ 125 \ from \ both \ sides:\\ y &= 75\end{align*}

**Anne rows at 125 yards per minute, and the river flows at 75 yards per minute.**

**Solving a Linear System by Multiplying Both Equations**

So what do we do if none of the coefficients match and none of them are simple multiples of each other? We do the same thing we do when we’re adding fractions whose denominators aren’t simple multiples of each other. Remember that when we add fractions, we have to find a **lowest common denominator**—that is, the lowest common multiple of the two denominators—and sometimes we have to rewrite not just one, but both fractions to get them to have a common denominator. Similarly, sometimes we have to multiply both equations by different constants in order to get one of the coefficients to match.

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 2060 miles in five days. Anne travels 880 miles from Norfolk, Virginia, and it takes her three days. If Anne pays $840 and Andrew pays $1845, what does I-Haul charge

a) *per day?*

b) *per mile traveled?*

**Solution**

First, we’ll set up our equations. Again we have 2 unknowns: the **daily rate** (we’ll call this \begin{align*}x\end{align*}), and the **per-mile rate** (we’ll call this \begin{align*}y\end{align*}).

Anne’s equation: \begin{align*}3x + 880y = 840\end{align*}

Andrew’s Equation: \begin{align*}5x + 2060y = 1845\end{align*}

We can’t just multiply a single equation by an integer number in order to arrive at matching coefficients. But if we look at the coefficients of \begin{align*}x\end{align*} (as they are easier to deal with than the coefficients of \begin{align*}y\end{align*}), we see that they both have a common multiple of 15 (in fact 15 is the ** lowest common multiple**). So we can multiply both equations.

Multiply the top equation by 5:

\begin{align*}15x + 4400y = 4200\end{align*}

Multiply the lower equation by 3:

\begin{align*}15x + 6180y = 5535\end{align*}

Subtract:

\begin{align*}& \qquad \qquad 15x + 4400y = 4200\\ & \underline{\quad \ - \ \ \ (15x + 6180y) = 5535\;\;\;\;\;\;}\\ & \qquad \qquad \quad \ \ -1780y = -1335\\ \\ & \text{Divide by}\ -1780: \ y = 0.75\end{align*}

Substitute this back into the top equation:

\begin{align*}3x + 880(0.75) &= 840 \\ 3x &= 180 \\ x &= 60\end{align*}

**I-Haul charges $60 per day plus $0.75 per mile.**

### Example

#### Example 1

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

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*}.

### Explore More

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*}
- \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*}

### Review (Answers)

To view the Review answers, open this PDF file and look for section 7.5.

### Notes/Highlights Having trouble? Report an issue.

Color | Highlighted Text | Notes | |
---|---|---|---|

Show More |

Term | Definition |
---|---|

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. |

Least Common Multiple |
The least common multiple of two numbers is the smallest number that is a multiple of both of the original numbers. |

### Image Attributions

Here you'll learn how to solve systems of linear equations in two variables by first multiplying and then eliminating one of the variables. You'll then solve real-world problems involving such systems.

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