# 7.4: Linear Systems with Addition or Subtraction

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

**Practice**Linear Systems with Addition or Subtraction

What if you were given a system of linear equations like \begin{align*}x + 4y = 7\end{align*} and \begin{align*}3x - 4y = -3\end{align*}? How could you solve for one of the variables by eliminating the other? After completing this Concept, you'll be able to solve a system of linear equations by elimination.

### Watch This

CK-12 Foundation: 0704S Solving Linear Systems by Elimination (H264)

### Guidance

In this lesson, we’ll see how to use simple addition and subtraction to simplify our system of equations to a single equation involving a single variable. Because we go from two unknowns (\begin{align*}x\end{align*} and \begin{align*}y\end{align*}) to a single unknown (either \begin{align*}x\end{align*} or \begin{align*}y\end{align*}), this method is often referred to by ** 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 A

*If one apple plus one banana costs $1.25 and one apple plus 2 bananas costs $2.00, how much does one banana cost? One apple?*

It shouldn’t take too long to discover that each banana costs $0.75. After all, the second purchase just contains 1 more banana than the first, and costs $0.75 more, so that one banana must cost $0.75.

Here’s what we get when we describe this situation with algebra:

\begin{align*}a + b &= 1.25\\ a + 2b &= 2.00\end{align*}

Now we can subtract the number of apples and bananas in the first equation from the number in the second equation, and also subtract the cost in the first equation from the cost in the second equation, to get the *difference* in cost that corresponds to the *difference* in items purchased.

\begin{align*}(a + 2b) - (a + b) = 2.00 - 1.25 \rightarrow b = 0.75\end{align*}

That gives us the cost of one banana. To find out how much one apple costs, we subtract $0.75 from the total cost of one apple and one banana.

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

So an apple costs 50 cents.

To solve systems using addition and subtraction, we’ll be using exactly this idea – by looking at the ** sum** or

**of the two equations we can determine a value for one of the unknowns.**

*difference*
**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 lets us combine two equations in such a way that the resulting equation has only one variable. We can then use simple algebra to solve for that variable. Then, if we need to, we can substitute the value we get for that variable back into either one of the original equations to solve for the other variable.

#### Example B

*Solve this system by addition:*

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

**Solution**

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:

\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 them together vertically, going down the columns. However, just like when you add units, tens and hundreds, you MUST be sure to keep the \begin{align*}x’\end{align*}s and \begin{align*}y’\end{align*}s in their own columns. You may also wish to use terms like \begin{align*}“0y”\end{align*} as a placeholder!

\begin{align*}& \ \quad \ \ 3x + 2y =11\\ &\underline{+ \ \ (5x - 2y)=13\;\;\;}\\ & \quad \ \ \ 8x + 0y = 24\end{align*}

Again we get \begin{align*}8x = 24\end{align*}, or \begin{align*}x = 3\end{align*}. To find a value for \begin{align*}y\end{align*}, we simply substitute our value for \begin{align*}x\end{align*} back in.

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

\begin{align*}5 \cdot 3 - 2y &= 13 && since \ 5 \times 3 = 15, \ we \ subtract \ 15 \ from \ both \ sides:\\ -2y &= -2 && divide \ by \ -2 \ to \ get:\\ y &= 1\end{align*}

The reason this method worked is that the \begin{align*}y-\end{align*}coefficients of the two equations were opposites of each other: 2 and -2. Because they were opposites, they canceled each other out when we added the two equations together, so our final equation had no \begin{align*}y-\end{align*}term in it and we could just solve it for \begin{align*}x\end{align*}.

**Solving Linear Systems Using Subtraction of Equations**

Another, very similar method for solving systems is subtraction. When the \begin{align*}x-\end{align*} or \begin{align*}y-\end{align*}coefficients in both equations are the same (including the sign) instead of being opposites, you can **subtract** one equation from the other.

If you look again at Example 3, you can see that the coefficient for \begin{align*}x\end{align*} in both equations is +1. Instead of adding the two equations together to get rid of the \begin{align*}y’\end{align*}s, you could have subtracted to get rid of the \begin{align*}x’\end{align*}s:

\begin{align*}&(x + y) - ( x - y) = 7 - 1.5 \Rightarrow 2y = 5.5 \Rightarrow y = 2.75\\ & \qquad \text{or}...\\ & \ \quad \ \ \ \ x + y = 7\\ & \underline{\;\; - \ \ (x - y) = -1.5\;\;\;\;\;}\\ & \quad \ 0x + 2y = 5.5\end{align*}

So again we get \begin{align*}y = 2.75\end{align*}, and we can plug that back in to determine \begin{align*}x\end{align*}.

The method of subtraction is just as straightforward as addition, so long as you remember the following:

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

#### Example C

*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?*

**Solution**

We have 2 types of coins, so let’s call the number of pennies \begin{align*}x\end{align*} and the number of nickels \begin{align*}y\end{align*}. The total value of all the pennies is just \begin{align*}x\end{align*}, since they are worth \begin{align*}1 \cancel{c}\end{align*} each. The total value of the nickels is \begin{align*}5y\end{align*}. We are given two key pieces of information to make our equations: the number of coins and their value in cents.

\begin{align*}& \# \ \text{of coins equation}: && x + y = 107 && (number \ of \ pennies) + (number \ of \ nickels)\\ & \text{value equation}: && x + 5y = 347 && pennies \ are \ worth \ 1 \cancel{c}, \ nickels \ are \ worth \ 5 \cancel{c}.\end{align*}

We’ll jump straight to subtracting the two equations:

\begin{align*}& \quad \quad \quad \ x + y = 107\\ & \underline{\;\;- \ \ (x + 5y) = -347\;\;}\\ & \quad \quad \quad \ -4y = -240\\ & \quad \quad \quad \quad \quad y = 60\end{align*}

Substituting this value back into the first equation:

\begin{align*}x + 60 &= 107 && subtract \ 60 \ from \ both \ sides:\\ x &= 47\end{align*}

**So Peter saw 47 pennies (worth 47 cents) and 60 nickels (worth $3.00) making a total of $3.47.**

Watch this video for help with the Examples above.

CK-12 Foundation: Linear Systems by Elimination

### Vocabulary

- 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.

### Guided Practice

*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, how fast is the current? How fast would Andrew travel in calm water?*

**Solution**

First we convert our problem into equations. We have two unknowns to solve for, so we’ll call the speed that Andrew paddles at \begin{align*}x\end{align*}, and the speed of the river \begin{align*}y\end{align*}. When traveling downstream, Andrew speed is boosted by the river current, so his total speed is his paddling speed *plus* the speed of the river \begin{align*}(x + y)\end{align*}. Traveling upstream, the river is working against him, so his total speed is his paddling speed *minus* the speed of the river \begin{align*}(x - y)\end{align*}.

Downstream Equation: \begin{align*}x + y = 7\end{align*}

Upstream Equation: \begin{align*}x - y = 1.5\end{align*}

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

\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 2:

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

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

\begin{align*}4.25 + y &= 7 && subtract \ 4.25 \ from \ both \ sides:\\ y &= 2.75\end{align*}

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

### Practice

- Solve the system: \begin{align*}3x + 4y = 2.5\!\\ 5x - 4y = 25.5\end{align*}
- Solve the system: \begin{align*}2x + -y = 10\!\\ 3x +y = -5\end{align*}
- Solve the system: \begin{align*}5x + 7y = -31\!\\ 5x - 9y = 17\end{align*}
- Solve the system: \begin{align*}3y - 4x = -33\!\\ 5x - 3y = 40.5\end{align*}
- Nadia and Peter visit the candy store. Nadia buys three candy bars and four fruit roll-ups for $2.84. 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 a candy bar and a fruit roll-up individually?
- 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.
- An airport taxi firm charges a pick-up fee, plus an additional per-mile fee for any rides taken. If a 12-mile journey costs $14.29 and a 17-mile journey costs $19.91, calculate:
- the pick-up fee
- the per-mile rate
- the cost of a seven mile trip

- 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 7-minute call costs $4.25 and a 12-minute call costs $5.50, find each rate.
- 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?
- 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?

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