Solving a System of Equations: Elimination

Section 6.3 Solving a System of Equations: Elimination

5

2 + 3

5 = 2 + 3

Example 6.3.1.

Let’s say we want to solve the following system of equations:

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{\begin{cases} 3 & = x + y \\ 5 & = 3 x - y \end{cases}

Remember that the symbol

=

means that whatever is on each side is exactly the same thing, even if it is written differently. As mentioned in the introduction, we can add the same thing to both sides, even if it looks different. So, let’s add 5 to both sides of the first equation, but we will write it as

3 x - y

on the right hand side. The equation

5 = 3 x - y

tells us that’s okay, because it means that

5

and

3 x - y

are the same thing:

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\begin{aligned} 3 + (5) & = x + y + (3 x - y) \\ 8 & = 4 x + y - y \\ 8 & = 4 x \end{aligned}

Notice what happened: the

y

’s canceled. So now, we can divide both sides by 4 to get

x = 2 .

Just like with substitution, we need to plug back into either of our starting equations to get

y :

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\begin{aligned} 3 & = x + y \\ 3 & = 2 + y \\ 1 & = y \end{aligned}

Now we have our answer of

(2, 1)

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Checkpoint 6.3.2.

Solve the following system of equations using elimination:

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{\begin{cases} 7 & = 2 x + 3 y \\ - 11 & = - 2 x + y \end{cases}

Answer.

(5, - 1)

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Solution.

We want to start by adding the equations together. In other words, we are going to add

7

and

- 11

on one side and then add

2 x + 3 y

and

- 2 x + y

on the other side. We know we can do this, because we are starting with things that are equal, so we are still adding the same thing to both sides, just written differently.

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\begin{aligned} 7 + (- 11) & = 2 x + 3 y + (- 2 x + y) \\ - 4 & = 2 x - 2 x + 4 y \\ - 4 & = 4 y \\ - 1 & = y \end{aligned}

Now that we have our answer for

y,

we just need to plug into one of the original equations to get

x :

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\begin{aligned} 7 & = 2 x + 3 y \\ 7 & = 2 x + 3 (- 1) \\ 7 & = 2 x - 3 \\ 10 & = 2 x \\ 5 & = x \end{aligned}

So we have our final answer:

(5, - 1) .

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Remember that when we talked about substitution, we saw two ways we could check our work:

After you find the value for one variable, plug it into both starting equations and make sure they give you the same value for the second variable

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After you have your final answer, plug the point into both starting equations and make sure it works for both

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You can still use both of these strategies to check your work with elimination.

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Now, notice why this strategy worked for the two problems above: when we added the equations together, one of the variables canceled out. It was eliminated. That only worked because that variable had the same number multiplied out front, but was negative in one equation and positive in the other. If the equation doesn’t have that situation to begin with, we might need to adjust one of the equations to make it work.

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Example 6.3.3.

Suppose we want to solve the following system of equations:

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{\begin{cases} - 6 & = 3 x + 4 y \\ - 2 & = x - 2 y \end{cases}

We have a couple of choices here:

If we multiply both sides of the second equation by $- 3,$ then we would be able to cancel the $x$ ’s

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If we multiply the second equation by $2,$ then we would be able to cancel the $y$ ’s.

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It doesn’t matter which one we choose, so let’s go with the first option for this example. When we multiply the second equation by

- 3,

we get

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6 = - 3 x + 6 y

Now, we add that to the first equation:

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\begin{aligned} - 6 + (6) & = 3 x + 4 y + (- 3 x + 6 y) \\ 0 & = 3 x - 3 x + 10 y \\ 0 & = 10 y \\ 0 & = y \end{aligned}

Now we will plug

y = 0

into one of the starting equations to get

x :

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\begin{aligned} - 2 & = x - 2 y \\ - 2 & = x - 2 (0) \\ - 2 & = x + 0 \\ - 2 & = x \end{aligned}

Therefore, we have our answer:

(- 2, 0) .

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Checkpoint 6.3.4.

Solve the following system using elimination.

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{\begin{cases} 4 & = x - 3 y \\ - 6 & = - 5 x + y \end{cases}