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# 7.3: Solving Linear Systems by Elimination through Addition or Subtraction

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Solve a linear system of equations using elimination by addition.
• Solve a linear system of equations using elimination by subtraction.
• Solve real-world problems using linear systems by elimination.

## Introduction

In this lesson, we will look at using simple addition and subtraction to simplify our system of equations to a single equation involving a single variable. Because we go from two unknowns (x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*}) to a single unknown (either x\begin{align*}x\end{align*} or y\begin{align*}y\end{align*}) this method is often referred to as solving by elimination. We eliminate one variable in order to make our equations solvable! To illustrate this idea, let’s look at the simple example of buying apples and bananas.

Example 1

If one apple plus one banana costs $1.25 and one apple plus two bananas costs$2.00, how much does it cost for one banana? One apple?

It shouldn’t take too long to discover that each banana costs 0.75. You can see this by looking at the difference between the two situations. Algebraically, using a\begin{align*}a\end{align*} and b\begin{align*}b\end{align*} as the cost for apples and bananas, we get the following equations. a+ba+2b=1.25=2.00\begin{align*}a + b & = 1.25 \\ a + 2b & = 2.00\end{align*} If you look at the difference between the two equations you see that the difference in items purchased is one banana, and the difference in money paid is 75 cents. So one banana costs 75 cents. (a+2b)(a+b)=2.001.25b=0.75\begin{align*}(a + 2b) - (a+b) = 2.00 - 1.25 \Rightarrow b = 0.75\end{align*} To find out how much one apple costs, we subtract0.75 from the cost of one apple and one banana. So an apple costs 50 cents.

a+0.75=1.25a=1.250.75a=0.50\begin{align*}a + 0.75 = 1.25 \Rightarrow a = 1.25 - 0.75 \Rightarrow a = 0.50\end{align*}

To solve systems using addition and subtraction, we will be using exactly this idea. By looking at the sum or difference of the two equations, we can determine a value for one of the unknowns.

## Solving Linear Systems Using Addition of Equations

Often considered the easiest and most powerful method of solving systems of equations, the addition (or elimination) method requires us to combine two equations in such a way that the resulting equation has only one variable. We can then use simple linear algebra methods of solving for that variable. If required, we can always substitute the value we get for that variable back in either one of the original equations to solve for the remaining unknown variable.

Example 2

3x+2y5x2y=11=13\begin{align*}3x+ 2y & = 11 \\ 5x- 2y & = 13\end{align*}

We will add everything on the left of the equals sign from both equations, and this will be equal to the sum of everything on the right.

(3x+2y)+(5x2y)=11+138x=24x=3\begin{align*}(3x + 2y) + (5x- 2y) = 11 + 13\Rightarrow 8x = 24 \Rightarrow x = 3\end{align*}

A simpler way to visualize this is to keep the equations as they appear above, and to add in columns. However, just like adding units tens and hundreds, you MUST keep x\begin{align*}x\end{align*}'s and y\begin{align*}y\end{align*}'s in their own columns. You may also wish to use terms like “0y\begin{align*}0y\end{align*}” as a placeholder!

3x+2y=11 +(3x2y)=138x+0y=24\begin{align*}& \quad \quad 3x + 2y = 11\\ & \ \underline{+ (3x - 2y) = 13\;\;}\\ & \qquad 8x + 0y = 24\end{align*}

Again we get 8x=24\begin{align*}8x = 24\end{align*} or x=3\begin{align*}x = 3\end{align*}.

To find a value for y\begin{align*}y\end{align*} we simply substitute our value for x\begin{align*}x\end{align*} back in.

Substitute x=3\begin{align*}x = 3\end{align*} into the second equation.

532y2yy=13=2=1Since 5×3=15, we subtract 15 from both sides.Divide by 2 to get the value for y.\begin{align*}5\cdot 3 - 2y & = 13 && \text{Since}\ 5 \times 3 = 15,\ \text{we subtract}\ 15\ \text{from both sides.}\\ -2y & = -2 && \text{Divide by}\ 2\ \text{to get the value for}\ y.\\ y & = 1 \end{align*}

The first example has a solution at x=3\begin{align*}x = 3\end{align*} and y=1\begin{align*}y =1\end{align*}. You should see that the method of addition works when the coefficients of one of the variables are opposites. In this case it is the coefficients of y\begin{align*}y\end{align*} that are opposites, being +2 in the first equation and -2 in the second.

There are other, similar, methods we can use when the coefficients are not opposites, but for now let’s look at another example that can be solved with the method of addition.

Example 3

Andrew is paddling his canoe down a fast moving river. Paddling downstream he travels at 7 miles per hour, relative to the river bank. Paddling upstream, he moves slower, traveling at 1.5 miles per hour. If he paddles equally hard in both directions, calculate, in miles per hour, the speed of the river and the speed Andrew would travel in calm water.

Step One First, we convert our problem into equations. We have two unknowns to solve for, so we will call the speed that Andrew paddles at x\begin{align*}x\end{align*}, and the speed of the river y\begin{align*}y\end{align*}. When traveling downstream, Andrew's speed is boosted by the river current, so his total speed is the canoe speed plus the speed of the river (x+y)\begin{align*}(x+y)\end{align*}. Upstream, his speed is hindered by the speed of the river. His speed upstream is (xy)\begin{align*}(x-y)\end{align*}.

Downstream EquationUpstream Equationx+yxy=7=1.5\begin{align*}\text{Downstream Equation} && x+y & =7 \\ \text{Upstream Equation} && x-y & =1.5\end{align*}

Step Two Next, we are going to eliminate one of the variables. If you look at the two equations, you can see that the coefficient of y\begin{align*}y\end{align*} is +1 in the first equation and -1 in the second. Clearly (+1)+(1)=0\begin{align*}(+1) + (-1) = 0\end{align*}, so this is the variable we will eliminate. To do this we add equation 1 to equation 2. We must be careful to collect like terms, and that everything on the left of the equals sign stays on the left, and everything on the right stays on the right:

(x+y)+(xy)=7+1.52x=8.5x=4.25\begin{align*}(x + y) + (x - y) = 7 + 1.5 \Rightarrow 2x = 8.5 \Rightarrow x = 4.25\end{align*}

Or, using the column method we used in example one.

x+y=7+ (xy)=1.52x+0y=8.5\begin{align*}& \qquad x + y = 7\\ & \underline{+ \ (x - y) = 1.5\;\;}\\ & \quad 2x + 0y = 8.5\end{align*}

Again you see we get 2x=8.5\begin{align*}2x = 8.5\end{align*}, or x=4.25\begin{align*}x = 4.25\end{align*}. To find a corresponding value for y\begin{align*}y\end{align*}, we plug our value for x\begin{align*}x\end{align*} into either equation and isolate our unknown. In this example, we’ll plug it into the first equation.

Substitute x=3\begin{align*}x = 3\end{align*} into the second equation:

4.25+yy=7=2.75Subtract 4.25 from both sides.\begin{align*}4.25 + y & = 7 && \text{Subtract}\ 4.25\ \text{from both sides.}\\ y & = 2.75\end{align*}

Solution

Andrew paddles at 4.25 miles per hour. The river moves at 2.75 miles per hour.

## Solving Linear Systems Using Subtraction of Equations

Another, very similar method for solving systems is subtraction. In this instance, you are looking to have identical coefficients for x\begin{align*}x\end{align*} or y\begin{align*}y\end{align*} (including the sign) and then subtract one equation from the other. If you look at Example one you can see that the coefficient for x\begin{align*}x\end{align*} in both equations is +1. You could have also used the method of subtraction.

(x+y)(xy)=200802y=120y=60\begin{align*}(x+y)-(x-y)=200 - 80 \Rightarrow 2y = 120 \Rightarrow y=60\end{align*}

or

x+y=200+ (xy)=800x+2y=120\begin{align*}& \qquad x + y = 200\\ & \underline{+ \ (x - y) = -80\;}\\ & \quad 0x + 2y = 120\end{align*}

So again we get y=60\begin{align*}y = 60\end{align*}, from which we can determine x\begin{align*}x\end{align*}. The method of subtraction looks equally straightforward, and it is so long as you remember the following:

1. Always put the equation you are subtracting in parentheses, and distribute the negative.
2. Don’t forget to subtract the numbers on the right hand side.
3. Always remember that subtracting a negative is the same as adding a positive.

Example 4

Peter examines the coins in the fountain at the mall. He counts 107 coins, all of which are either pennies or nickels. The total value of the coins is 3.47. How many of each coin did he see? We have two types of coins. Let’s call the number of pennies x\begin{align*}x\end{align*} and the number of nickels y\begin{align*}y\end{align*}. The total value of pennies is just x\begin{align*}x\end{align*}, since they are worth one cent each. The total value of nickels is 5y\begin{align*}5y\end{align*}. We are given two key pieces of information to make our equations. The number of coins and their value. \begin{align*}\text{Number of Coins Equation} && x + y = 107 && \text{(number of pennies)} + \text{(number of nickels)}\\ \text{The Value Equation:} && x + 5y = 347 && \text{pennies are worth}\ 1c, \text{nickels are worth}\ 5c.\end{align*} We will jump straight to the subtraction of the two equations. \begin{align*} & \qquad \ x + y \ = \ 107\\ & \underline{+ \ (x + 5y) = -347\;}\\ & \qquad \quad \ \ 4y = -240\end{align*} Let's substitute this value back into the first equation. \begin{align*}x+60 & = 107 && \text{Subtract}\ 60\ \text{from both sides.}\\ x & = 47 \end{align*} So Peter saw 47 pennies (worth 47 cents) and 60 nickels (worth3.00) for a total of 3.47. We have now learned three techniques for solving systems of equations. 1. Graphing 2. Substitution 3. Elimination You should be starting to gain an understanding of which method to use when given a particular problem. For example, graphing is a good technique for seeing what the equations are doing, and when one service is less expensive than another. Graphing alone may not be ideal when an exact numerical solution is needed. Similarly, substitution is a good technique when one of the coefficients in your equation is +1 or -1. Addition or subtraction is ideal when the coefficient of one of the variables matches the coefficient of the same variable in the other equation. In the next lesson, we will learn the last technique for solving systems of equations exactly, when none of the coefficients match and the coefficient is not one. Multimedia Link The following video contains three examples of solving systems of equations using multiplication and addition and subtraction as well as multiplication (which is the next topic). Khan Academy Systems of Equations (9:57) . Note that the narrator is not always careful about showing his work, and you should try to be neater in your mathematical writing. ## Review Questions 1. Solve the system: \begin{align*}3x + 4y = 2.5\!\\ 5x-4y = 25.5\end{align*} 2. Solve the system \begin{align*}5x + 7y = -31\!\\ 5x-9y = 17\end{align*} 3. Solve the system \begin{align*}3y-4x = -33\!\\ 5x-3y = 40.5\end{align*} 4. Nadia and Peter visit the candy store. Nadia buys three candy bars and four fruit roll-ups for2.84. Peter also buys three candy bars, but can only afford one additional fruit roll-up. His purchase costs $1.79. What is the cost of each candy bar and each fruit roll-up? 5. A small plane flies from Los Angeles to Denver with a tail wind (the wind blows in the same direction as the plane) and an air-traffic controller reads its ground-speed (speed measured relative to the ground) at 275 miles per hour. Another, identical plane, moving in the opposite direction has a ground-speed of 227 miles per hour. Assuming both planes are flying with identical air-speeds, calculate the speed of the wind. 6. An airport taxi firm charges a pick-up fee, plus an additional per-mile fee for any rides taken. If a 12 miles journey costs$14.29 and a 17 miles journey costs $19.91, calculate: 1. the pick-up fee 2. the per-mile rate 3. the cost of a seven mile trip 7. Calls from a call-box are charged per minute at one rate for the first five minutes, then a different rate for each additional minute. If a seven minute call costs$4.25 and a 12 minute call costs $5.50, find each rate. 8. A plumber and a builder were employed to fit a new bath, each working a different number of hours. The plumber earns$35 per hour, and the builder earns $28 per hour. Together they were paid$330.75, but the plumber earned $106.75 more than the builder. How many hours did each work? 9. Paul has a part time job selling computers at a local electronics store. He earns a fixed hourly wage, but can earn a bonus by selling warranties for the computers he sells. He works 20 hours per week. In his first week, he sold eight warranties and earned$220. In his second week, he managed to sell 13 warranties and earned 280. What is Paul’s hourly rate, and how much extra does he get for selling each warranty? ## Review Answers 1. \begin{align*}x = 3.5, y = -2\end{align*} 2. \begin{align*}x= -2, y = -3\end{align*} 3. \begin{align*}x = 7.5, y = -1 \end{align*} 4. Candy bars cost 48 cents each and fruit roll-ups cost 35 cents each. 5. The wind speed is 24 mph 1..80
2. $1.12 3.$8.64
6. 75 cents per minute for the first 5 mins, 25 cents per minute additional
7. The plumber works 6.25 hours, the builder works 4 hours
8. Paul earns a base of \$7.00 per hour

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