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# 7.4: Linear Systems with Addition or Subtraction

Difficulty Level: At Grade Created by: CK-12
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Practice Linear Systems with Addition or Subtraction
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What if you were given a system of linear equations like $x + 4y = 7$ and $3x - 4y = -3$ ? 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.

### 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 ( $x$ and $y$ ) to a single unknown (either $x$ or $y$ ), 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: $a + b &= 1.25\\a + 2b &= 2.00$ 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. $(a + 2b) - (a + b) = 2.00 - 1.25 \rightarrow b = 0.75$ 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.

$a + 0.75 = 1.25 \rightarrow a = 1.25 - 0.75 \rightarrow a = 0.50$

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 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 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

$3x + 2y &= 11\\5x - 2y &= 13$

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:

$(3x + 2y) + (5x - 2y) = 11 + 13 \rightarrow 8x = 24 \rightarrow x = 3$

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 $x$ 's and $y$ 's in their own columns. You may also wish to use terms like " $0y$ " as a placeholder!

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

Again we get $8x = 24$ , or $x = 3$ . To find a value for $y$ , we simply substitute our value for $x$ back in.

Substitute $x = 3$ into the second equation:

$5 \cdot 3 - 2y &= 13 && since \ 5 \times 3 = 15, \ we \ subtract \ 15 \ from \ both \ sides\\-2y &= -2 && divide \ by \ -2\\y &= 1$

The reason this method worked is that the $y-$ 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 $y-$ term in it and we could just solve it for $x$ .

Solving Linear Systems Using Subtraction of Equations

Another, very similar method for solving systems is subtraction. When the $x-$ or $y-$ 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 $x$ in both equations is +1. Instead of adding the two equations together to get rid of the $y'$ s, you could have subtracted to get rid of the $x'$ s:

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

So again we get $y = 2.75$ , and we can plug that back in to determine $x$ .

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.

## Date Created:

Aug 13, 2012

Oct 28, 2014
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