ORCCA Elimination

Section 4.3 Elimination

Objectives: PCC Course Content and Outcome Guide

In Section 1, we used graphing to solve a system of two linear equations. In Section 2, we learned the substitution technique. In this section, we learn another technique for solving a system of linear equations called “elimination” or “the addition method”.

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Figure 4.3.1. Alternative Video Lesson

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Subsection 4.3.1 Solving Systems of Equations by Elimination

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Example 4.3.2.

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Alicia has

$ 1000

to send to her two grandchildren Bobbie and Cedric. Since Bobbie is entering college and has some textbook expenses, Alicia wants to send Bobbie

$ 200

more than she sends Cedric. How much money should she give to each grandchild?

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You may have a good way to answer this quickly, but we will use this example to demonstrate a new technique for solving a system of two linear equations.

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Let

B

be the dollar amount that she sends to Bobbie, and

C

be the dollar amount that she sends to Cedric. (Note that we begin the process of solving a word problem by clearly defining the variables, including their units.) Since the total she has to give is

$ 1000,

we can write

B + C = 1000 .

And since she wants to give

$ 200

more to Bobbie, we can write

B - C = 200 .

So we have the system of equations:

\begin{array}{r} {\begin{aligned} B + C & = 1000 \\ B - C & = 200 \end{aligned} \end{array}

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We could solve this system using graphing or substitution, but here is a new method. If we add together the left sides from the two equations, it should equal the sum of the right sides:

\begin{aligned} \begin{matrix} B + C \\ \underset{―}{+ B - C} \\ 2 B \end{matrix} & \begin{array}{r} = \end{array} \begin{array}{r} 1000 \\ \underset{―}{+ 200} \\ 1200 \end{array} \end{aligned}

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Note that the variable

C

is eliminated. This happened because the “

+ C

” and the “

- C

” perfectly cancel each other out when they are added. With only one variable left, it doesn’t take much effort to finish solving for

B :

B = 600 .

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To finish solving this system of equations, we still need the value of

C .

One easy way to find

C

is to substitute our value for

B

into one of the original equations:

\begin{aligned} B + C & = 1000 \\ 600 + C & = 1000 \\ C & = 400 \end{aligned}

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To check our work, substitute

B = 600

and

C = 400

into the original equations:

\begin{aligned} B + C & = 1000 & B - C & = 200 \\ 600 + 400 & \overset{✓}{=} 1000 & 600 - 400 & \overset{✓}{=} 200 \end{aligned}

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This confirms that our solution is correct. In summary, Alicia should give

$ 600

to Bobbie and

$ 400

to Cedric.

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This method for solving a system of equations worked because

C

and

- C

add to zero. Once the

C

-terms were eliminated we were able to solve for

B .

This method is called the elimination method or the addition method, because we add the corresponding sides from the two equations and that eliminates a variable.

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Most of the time, just adding the sides together does not eliminate one of the variables. We can still use this method, but it needs a little setup work first. Let’s look at an example where we need to adjust one of the equations.

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Example 4.3.3. Scaling One Equation.

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Solve the system of equations using the elimination method.

\begin{array}{r} {\begin{aligned} 3 x & - & 4 y & = & 2 \\ 5 x & + & 8 y & = & 18 \end{aligned} \end{array}

Explanation.

We want to see whether it will be easier to eliminate

x

y .

We see that the coefficients of

x

in each equation are

3

and

5,

and the coefficients of

y

are

- 4

and

8 .

Because

8

is a multiple of

4

and the coefficients already have opposite signs, the

y

variable will be easier to eliminate.

To eliminate the

y

terms, first multiply each side of the first equation by

2 .

Why? Because then the

y

-terms will be

8 y

and

- 8 y,

so they can be added to eliminate

y .

\begin{aligned} {\begin{aligned} 2 \cdot (3 x & - & 4 y) & = & 2 \cdot (2) \\ 5 x & + & 8 y & = & 18 \end{aligned} \\ {\begin{aligned} 6 x & - & 8 y & = & 4 \\ 5 x & + & 8 y & = & 18 \end{aligned} \end{aligned}

We now have an equivalent system of equations where the

y

-terms can be eliminated:

\begin{aligned} \begin{matrix} 6 x - 8 y \\ \underset{―}{+ 5 x + 8 y} \\ 11 x \end{matrix} & \begin{array}{r} = \end{array} \begin{array}{r} 4 \\ \underset{―}{+ 18} \\ 22 \end{array} \end{aligned}

So we have

x = 2 .

To solve for

y,

one option is to substitute

2

in for

x

into either of the original equations. We use the first original equation,

3 x - 4 y = 2 :

\begin{aligned} 3 x - 4 y & = 2 \\ 3 (2) - 4 y & = 2 \\ 6 - 4 y & = 2 \\ - 4 y & = - 4 \\ y & = 1 \end{aligned}

We think the solution is

x = 2

and

y = 1 .

We can check this using both of the original equations:

\begin{aligned} 5 x + 8 y & = 18 & 3 x - 4 y & = 2 \\ 5 (2) + 8 (1) & \overset{?}{=} 18 & 3 (2) - 4 (1) & \overset{?}{=} 2 \\ 10 + 8 & \overset{✓}{=} 18 & 6 - 4 & \overset{✓}{=} 2 \end{aligned}

So the solution to this system is

(2, 1)

and the solution set is

{(2, 1)} .

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Checkpoint 4.3.4.

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Use elimination to solve the system.

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\begin{array}{r} {\begin{aligned} 5 x & + & 6 y & = & - 7 \\ 4 x & + & 2 y & = & - 1 \end{aligned} \end{array}

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(a)

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What number could you multiply by each term in the second equation that would be helpful?

Explanation.

If we multiply the second equation through by

- 3

then the

y

-terms will be

6 y

and

- 6 y,

and adding will eliminate

y .

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(b)

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After multiplying the terms in the second equation by

- 3

and adding the two equations together, solve for

x .

Explanation.

\begin{aligned} {\begin{aligned} 5 x & + & 6 y & = & - 7 \\ - 3 \cdot (4 x & + & 2 y) & = & - 3 \cdot (- 1) \end{aligned} \\ {\underset{―}{\begin{aligned} 5 x & + & 6 y & = & - 7 \\ - 12 x & - & 6 y & = & 3 \end{aligned}} \\ \begin{aligned} - 7 x & = - 4 \\ x & = \frac{4}{7} \end{aligned} \end{aligned}

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(c)

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Now solve for

y

as well, by substituting

\frac{4}{7}

in for

x

in one of the original equations.

Explanation.

\begin{aligned} 4 x + 2 y & = - 1 \\ 4 (\frac{4}{7}) + 2 y & = - 1 \\ \frac{16}{7} + 2 y & = - 1 \\ 7 \cdot (\frac{16}{7} + 2 y) & = 7 \cdot (- 1) \\ 16 + 14 y & = - 7 \\ 14 y & = - 23 \\ y & = - \frac{23}{14} \end{aligned}

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Here’s an example where we have to scale both equations.

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Example 4.3.5. Scaling Both Equations.

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Solve the system of equations using the elimination method.

\begin{array}{r} {\begin{aligned} 2 x & + & 3 y & = & 10 \\ - 3 x & + & 5 y & = & - 15 \end{aligned} \end{array}

Explanation.

Considering the coefficients of

x

(

2

and

- 3

) and the coefficients of

y

(

3

and

5

) we see that we cannot eliminate the

x

or the

y

variable by scaling a single equation. There’s just no place where one coefficient divides the corresponding coefficient. But we can scale both equations.

The

x

-terms already have opposite signs, so we choose to eliminate

x .

The least common multiple of

2

and

3

6,

so we can target turning the terms into

6 x

and

- 6 x .

We can scale the first equation by

3

and the second equation by

2

so that the equations have terms

6 x

and

- 6 x,

which will cancel when added.

\begin{aligned} {\begin{aligned} 3 \cdot (2 x & + & 3 y) & = & 3 \cdot (10) \\ 2 \cdot (- 3 x & + & 5 y) & = & 2 \cdot (- 15) \end{aligned} \\ {\underset{―}{\begin{aligned} 6 x & + & 9 y & = & 30 \\ - 6 x & + & 10 y & = & - 30 \end{aligned}} \\ \begin{aligned} 19 y & = 0 \\ y & = 0 \end{aligned} \end{aligned}

To solve for

x,

we’ll replace

y

with

0

2 x + 3 y = 10 :

\begin{aligned} 2 x + 3 y & = 10 \\ 2 x + 3 (0) & = 10 \\ 2 x & = 10 \\ x & = 5 \end{aligned}

We’ll check the system using

x = 5

and

y = 0

in each of the original equations:

\begin{aligned} 2 x + 3 y & = 10 & - 3 x + 5 y & = - 15 \\ 2 (5) + 3 (0) & \overset{?}{=} 10 & - 3 (5) + 5 (0) & \overset{?}{=} - 15 \\ 10 + 0 & \overset{✓}{=} 10 & - 15 + 0 & \overset{✓}{=} - 15 \end{aligned}

So the system’s solution is

(5, 0)

and the solution set is

{(5, 0)} .

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Checkpoint 4.3.6.

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Try a similar exercise.

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Use elimination to solve the system.

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\begin{array}{r} {\begin{aligned} 3 x & + & 4 y & = & - 26 \\ 5 x & + & 5 y & = & - 40 \end{aligned} \end{array}

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(a)

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What numbers could you multiply each term in the first and second equations by that would be helpful?

Explanation.

If we multiply the first equation through by

5

and the second equation through by

- 3

then the

x

-terms will be

15 x

and

- 15 x,

and adding will eliminate

x .

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(b)

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After multiplying the terms in the first by

5

and the terms in the second equation by

- 3,

and adding the two equations together, solve for

y .

Explanation.

\begin{aligned} {\begin{aligned} 5 \cdot (3 x & + & 4 y) & = & 5 \cdot (- 26) \\ - 3 \cdot (5 x & + & 5 y) & = & - 3 \cdot (- 40) \end{aligned} \\ {\underset{―}{\begin{aligned} 15 x & + & 20 y & = & - 130 \\ - 15 x & - & 15 y & = & 120 \end{aligned}} \\ \begin{aligned} 5 y & = - 10 \\ y & = - 2 \end{aligned} \end{aligned}

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(c)

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Now solve for

x

as well, by substituting

- 2

in for

y

in one of the original equations.

Explanation.

\begin{aligned} 3 x + 4 y & = - 26 \\ 3 x + 4 (- 2) & = - 26 \\ 3 x - 8 & = - 26 \\ 3 x & = - 18 \\ x & = - 6 \end{aligned}

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Example 4.3.7. Meal Planning.

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Javed is on a meal plan where his breakfast needs to have

600

calories and

20

grams of fat. A small avocado contains

300

calories and

30

grams of fat. He has bagels with

400

calories and

8

grams of fat. Write and solve a system of equations to determine how much bagel and how much avocado Javed could eat to meet his target calories and fat.

Explanation.

To write this system of equations, we first need to define our variables. Let

A

be the number of avocados consumed and let

B

be the number of bagels consumed. (In the end, both

A

and

B

might be fractions.) For our first equation, we can count calories contributed from eating

A

avocados and

B

bagels:

(300 \frac{calories}{avocado}) (A avocados) + (400 \frac{calories}{bagels}) (B bagel) = 600 calories

Or, without the units:

300 A + 400 B = 600 .

For a second equation, we can count the grams of fat:

(30 \frac{g fat}{avocado}) (A avocados) + (8 \frac{g fat}{bagels}) (B bagel) = 20 g fat

Or, without the units:

30 A + 8 B = 20 .

So the system of equations is:

\begin{array}{r} {\begin{aligned} 300 A & + & 400 B & = & 600 \\ 30 A & + & 8 B & = & 20 \end{aligned} \end{array}

Since none of the coefficients are equal to

1,

it may be easier to use the elimination method to solve this system than it would be to use substitution. Looking at the terms

300 A

and

30 A,

we can eliminate the

A

variable if we multiply the second equation by

- 10

to get the term

- 300 A :

\begin{aligned} {\begin{aligned} 300 A & + & 400 B & = 600 \\ - 10 \cdot (30 A & + & 8 B) & = - 10 \cdot (20) \end{aligned} \\ {\begin{aligned} 300 A & + & 400 B & = & 600 \\ - 300 A & - & 80 B & = & - 200 \end{aligned} \end{aligned}

When we add the corresponding sides from the two equations together we have:

\begin{aligned} 320 B & = 400 \\ \frac{320 B}{320} & = \frac{400}{320} \\ B & = \frac{5}{4} \end{aligned}

Now we know that Javed should eat

\frac{5}{4}

of a bagel (in other words, one and one-quarter bagels). To determine how much avocado he should eat, we can substitute

\frac{5}{4}

in for

B

in either of our original equations.

\begin{aligned} 300 A + 400 B & = 600 \\ 300 A + 400 (\frac{5}{4}) & = 600 \\ 300 A + 500 & = 600 \\ 300 A & = 100 \\ \frac{300 A}{300} & = \frac{100}{300} \\ A & = \frac{1}{3} \end{aligned}

We think the solution is

A = \frac{1}{3},

B = \frac{5}{4} .

To check this, see if these numbers work as solutions in the original equations:

\begin{aligned} 300 A + 400 B & = 600 & 30 A + 8 B & = 20 \\ 300 (\frac{1}{3}) + 400 (\frac{5}{4}) & \overset{?}{=} 600 & 30 (\frac{1}{3}) + 8 (\frac{5}{4}) & \overset{?}{=} 20 \\ 100 + 500 & \overset{✓}{=} 600 & 10 + 10 & \overset{✓}{=} 20 \end{aligned}

In summary, Javed can eat

\frac{5}{4}

of a bagel (so one and one-quarter bagel) and

\frac{1}{3}

of an avocado in order to consume exactly

600

calories and

20

grams of fat.

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For summary reference, here is the general procedure.

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Process 4.3.8. Solving Systems of Equations by Elimination.

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To solve a system of equations by elimination,

Algebraically manipulate both equations into standard form if they are not already that way.
Scale one or both of the equations to force one of the variables to have equal but opposite coefficients in the two equations.
Add the corresponding sides of the two equations together, which should have the effect that one variable is eliminated entirely.
Solve the resulting equation for the one remaining variable.
Substitute that value into either of the original equations to find the other variable.
Verify your solution using the original two equations.

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Subsection 4.3.2 Solving Special Systems of Equations with Elimination

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Remember the two special cases for a system of two linear equations? In Subsection 4.1.3, we learned how such systems have lines that are parallel when graphed. If you are using the substitution method, we learned in Subsection 4.2.3 that after you make the substitution and simplify, you get an equation that is either outright true or outright false.

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With these systems, what happens when you use the elimination method? Let’s see.

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Example 4.3.9. A System with Infinitely Many Solutions.

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Solve the system of equations using the elimination method.

\begin{array}{r} {\begin{aligned} 3 x & + & 4 y & = & 5 \\ 6 x & + & 8 y & = & 10 \end{aligned} \end{array}

Explanation.

To eliminate the

x

-terms, we multiply each term in the first equation by

- 2,

and we have:

\begin{aligned} {\begin{aligned} - 2 \cdot (3 x & + & 4 y) & = & - 2 \cdot 5 \\ 6 x & + & 8 y & = & 10 \end{aligned} \\ {\begin{aligned} - 6 x & - & 8 y & = & - 10 \\ 6 x & + & 8 y & = & 10 \end{aligned} \end{aligned}

You might notice that the equations look very similar. Adding the respective sides of the equation, we have:

0 = 0

Not just one, but both of the variables have been eliminated. This is not giving us any restrictions on what

x

- and

y

-values would solve this system. The two equations were essentially the same, after scaling them. So they represent the same line, and the solution set contains all pairs

(x, y)

that lie on that line. We can write the solution set as

{(x, y) ∣ 3 x + 4 y = 5} .

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Example 4.3.10. A System with No Solution.

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Solve the system of equations using the elimination method.

\begin{array}{r} {\begin{aligned} 10 x & + & 6 y & = & 9 \\ 25 x & + & 15 y & = & 4 \end{aligned} \end{array}

Explanation.

To eliminate the

x

-terms, we will scale the first equation by

- 5

and the second by

2 :

\begin{aligned} {\begin{aligned} - 5 \cdot (10 x & + & 6 y) & = & - 5 \cdot (9) \\ 2 \cdot (25 x & + & 15 y) & = & 2 \cdot (4) \end{aligned} \\ {\begin{aligned} - 50 x & - & 30 y & = & - 45 \\ 50 x & + & 30 y & = & 8 \end{aligned} \end{aligned}

Adding the respective sides of the equation, we have:

0 = - 37

Again, not just one, but both of the variables have been eliminated. In this case, the statement

0 = - 37

is outright false no matter what

x

and

y

are. So the system has no solution.

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Subsection 4.3.3 Substitution versus Elimination

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In every example so far from this section, both equations were in standard form,

A x + B y = C .

And all of the coefficients were integers. If none of the coefficients are equal to

1

then it is usually easier to use the elimination method, because otherwise you will probably have some fraction arithmetic to do in the middle of the substitution method. If there is a coefficient of

1,

then it’s a matter of preference. When one of the coefficients is

1,

it’s not so bad to isolate that variable and make a substitution.

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Example 4.3.11.

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A college used to have a north campus with

6000

students and a south campus with

15,000

students. The percentage of students at the north campus who self-identify as LGBTQ was three times the percentage at the south campus. After the merge,

5.5 %

of students identify as LGBTQ. What percentage of students on each campus identified as LGBTQ before the merge?

Explanation.

We will define

N

as the percentage (as a decimal) of students at the north campus and

S

as the percentage (as a decimal) of students at the south campus that identified as LGBTQ. Since the percentage of students at the north campus was three times the percentage at the south campus, we have one equation:

N = 3 S

For a second equation, we will count LGBTQ students at the various campuses. At the north campus, multiply the population,

6000,

by the percentage

N

to get

6000 N .

This represents be the actual count of LGBTQ students at that campus. Similarly, the south campus had

15000 S

LGBTQ students, and the combined school has

21000 (0.055) = 1155 .

When we combine the two campuses, we have:

6000 N + 15000 S = 1155

So we have a system of two equations:

{\begin{aligned} N & = 3 S \\ 6000 N + 15000 S & = 1155 \end{aligned}

Because the first equation already has

N

isolated, this is a good time to not use the elimination method. We can directly substitute

3 S

in for

N

in our second equation and solve for

S :

\begin{aligned} 6000 N + 15000 S & = 1155 \\ 6000 (3 S) + 15000 S & = 1155 \\ 18000 S + 15000 S & = 1155 \\ 33000 S & = 1155 \\ \frac{33000 S}{33000} & = \frac{1155}{33000} \\ S & = 0.035 \end{aligned}

And then we can determine

N

using the first equation:

\begin{aligned} N & = 3 S \\ N & = 3 (0.035) \\ N & = 0.105 \end{aligned}

So before the merge,

10.5 %

of the north campus students self-identified as LGBTQ and

3.5 %

of the south campus students self-identified as LGBTQ.

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If you need to solve a system, and one of the equations is not in standard form, substitution may be easier. But you also may find it easier to convert the equations into standard form and use elimination. Additionally, if the system’s coefficients are fractions or decimals, it can help to take one step where you clear denominators and/or decimal points and then decide which technique you would like to use.

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Example 4.3.12.

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Solve the system of equations using the method of your choice.

\begin{array}{r} {\begin{aligned} - \frac{1}{3} y & = \frac{1}{15} x + \frac{1}{5} \\ \frac{5}{2} x - y & = 6 \end{aligned} \end{array}

Explanation.

First, we can clear the denominators by using the least common multiple of the denominators in each equation, similarly to in Section 2.3. We have:

\begin{aligned} {\begin{aligned} 15 \cdot - \frac{1}{3} y & = 15 \cdot (\frac{1}{15} x + \frac{1}{5}) \\ 2 \cdot (\frac{5}{2} x - y) & = 2 \cdot (6) \end{aligned} \\ {\begin{aligned} - 5 y & = x + 3 \\ 5 x - 2 y & = 12 \end{aligned} \end{aligned}

We could convert the first equation into standard form by subtracting

x

from both sides, and then use elimination.

\begin{aligned} {\begin{aligned} - x - 5 y & = 3 \\ 5 x - 2 y & = 12 \end{aligned} \end{aligned}

Alternatively, since the

x

-variable in the first equation has coefficient

1,

you might prefer the substitution method. Isolating

x

gives us

x = - 5 y - 3,

and we could substitute that in to the second equation. It’s really up to you. As an exercise, try it both ways. Hopefully you get the same solution either way, and you can reflect on which method feels more comfortable to you.

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This next example is an application exercise with decimal coefficients, and it demonstrates how you might clear decimals, similarly to clearing denominators.

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Example 4.3.13.

🔗

A penny is made from copper and zinc. A chemistry reference says copper has a density of 9 ^g⁄_cm³ and zinc has a density of 7.1 ^g⁄_cm³. A penny’s mass is 2.5 g and its volume is 0.35 cm³. How many cm³ each of copper and zinc go into one penny?

Explanation.

Let

C

be the volume of copper and

Z

be the volume of zinc in one penny, both measured in cm³. Since the total volume is 0.35 cm³, one equation is:

(C {cm}^{3}) + (Z {cm}^{3}) = 0.35 {cm}^{3}

Or without units:

C + Z = 0.35 .

For a second equation, we will examine the masses of copper and zinc. Since copper has a density of 9 ^g⁄_cm³ and we are using

C

to represent the volume of copper, the mass of copper in one penny is

9 C .

Similarly, the mass of zinc is

7.1 Z .

Since the total mass is 2.5 g, we have the equation:

(9 \frac{g}{{cm}^{3}}) (C {cm}^{3}) + (7.1 \frac{g}{{cm}^{3}}) (Z {cm}^{3}) = 2.5 g

Or without units:

9 C + 7.1 Z = 2.5 .

So we have a system of equations:

\begin{array}{r} {\begin{aligned} C & + & Z & = & 0.35 \\ 9 C & + & 7.1 Z & = & 2.5 \end{aligned} \end{array}

We can eliminate the decimals by multiplying by the appropriate power of

10 .

If each term in the first equation is multiplied by

100,

and each term in the second equation is multiplied by

10,

there are no more decimal points.

\begin{aligned} {\begin{aligned} 100 \cdot (C & + & Z) & = & 100 \cdot (0.35) \\ 10 \cdot (9 C & + & 7.1 Z) & = & 10 \cdot (2.5) \end{aligned} \\ {\begin{aligned} 100 C & + 100 Z & = & 35 \\ 90 C & + 71 Z & = & 25 \end{aligned} \end{aligned}

Now to set up elimination, scale each equation again to eliminate

C :

\begin{aligned} {\begin{aligned} 9 \cdot (100 C & + & 100 Z) & = & 9 \cdot (35) \\ - 10 \cdot (90 C & + & 71 Z) & = & - 10 \cdot (25) \end{aligned} \\ {\begin{aligned} 900 C & + 900 Z & = & 315 \\ - 900 C & - 710 Z & = & - 250 \end{aligned} \end{aligned}

Adding the corresponding sides from the two equations gives

190 Z = 65,

from which we find

Z = \frac{65}{190} \approx 0.342 .

So there is about 0.342 cm³ of zinc in a penny.

To solve for

C,

we can use one of the original equations:

\begin{aligned} C + Z & = 0.35 \\ C + 0.342 & \approx 0.35 \\ C & \approx 0.008 \end{aligned}

Therefore there is about 0.342 cm³ of zinc and 0.008 cm³ of copper in a penny.

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If a variable in a system is already isolated in one of the equations, or has a coefficient of

1,

consider using the substitution method. If both equations are in standard form or none of the coefficients are equal to

1,

we suggest using the elimination method. Either way, if you have fraction or decimal coefficients, it may help to scale your equations so that only integer coefficients remain.

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Reading Questions 4.3.4 Reading Questions

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1.

🔗

What is another name for the “elimination method”?

🔗

2.

🔗

To use the elimination method, usually the first step is to at least one equation.

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3.

🔗

Describe a good situation to use the substitution method instead of the elimination method for solving a system of two linear equations in two variables.

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