Solve Systems of Equations Using Elimination Method

The aim of this lesson is to show a method to solve a system of linear equations by adding or subtracting the equations.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Challenge

How Many Toys Are There in Each Box?

Maya is researching a toy factory that currently produces robot action figures and car toys.

These toys are sold to stores in boxes. The total weight of the content of each box is of

20

kilograms. Each robot weighs

\frac{4}{5}

kilogram and each car weighs

\frac{1}{1 0}

kilogram. If

x

is the number of robots and

y

the number of cars in a box, the given information can be expressed by a linear equation.

\frac{4}{5} x + \frac{1}{1 0} y = 20

Also, each robot and each car are sold to the stores for

$ 15

and

$ 5,

respectively. The factory makes

$ 500

for each box sold to a store. This information can also be written as an equation.

15 x + 5 y = 500

These equations can be combined to form a system of linear equations.

{\frac{4}{5} x + \frac{1}{1 0} y = 20 15 x + 5 y = 500

Solve this system of equations without graphing. How many robots and cars are there in each box?

Discussion

Rewriting Equations

Sometimes the equations of a system can be rewritten in a way that is easier to work with. As an example, consider the following system of linear equations.

{x + 2 y = 3 \frac{1}{2} x - y = - \frac{1}{2} (I) (II)

In Equation (II), the coefficient of the variable

x

is a fraction. It would be easier to solve the system if all the coefficients were integers. It is possible to rewrite the system using equivalent systems.

Rule

Equivalent Systems

In a system of equations, an equivalent system can be created by replacing one equation with the sum of two or more equations in the system or by replacing an equation with a multiple of itself. An equation can be also replaced by the sum of that equation and a multiple of another equation in the system.

Proof

The statement will be proved for linear equations. For non-linear equations, the proof is similar. Consider a system of linear equations.

{a x + b y = c d x + e y = f (I) (II)

Let

(x_{1}, y_{1})

be a solution to the system. Therefore, this ordered pair satisfies both equations simultaneously.

{a x_{1} + b y_{1} = c ✓ d x_{1} + e y_{1} = f ✓

Consider now the system formed by the following two equations.

Equation (I)
Equation (II) plus a multiple of Equation (I)

The system described above is shown below.

{a x + b y = c d x + e y + α (a x + b y) = f + α c

It is already known that

(x_{1}, y_{1})

is a solution to the first equation of this new system. It needs to be verified that the ordered pair is also a solution to the second equation.

{a x_{1} + b y_{1} = c ✓ d x_{1} + e y_{1} + α (a x_{1} + b y_{1}) = ? f + α c

Since

(x_{1}, y_{1})

is a solution to the original system, it is known that

a x_{1} + b y_{1} = c

and that

d x_{1} + e y_{1} = f .

These two expressions can be substituted into the second equation above.

d x_{1} + e y_{1} + α (a x_{1} + b y_{1}) = ? f + α c

SubstituteII

$a x_{1} + b y_{1} = c$ , $d x_{1} + e y_{1} = f$

f + α c = f + α c ✓

Therefore,

(x_{1}, y_{1})

is also a solution to the second equation of the new system. This means that this ordered pair is a solution to the system formed by the first equation of the original system and the sum of the first equation and a multiple of the second equation.

{a x_{1} + b y_{1} = c ✓ d x_{1} + e y_{1} + α (a x_{1} + b y_{1}) = f + α c ✓

Equivalent systems can be used to solve systems of equations. One method to do this is called the Elimination Method.

Discussion

Elimination Method

Given a system of two equations in two variables, replacing one equation with the sum of that equation and a multiple of the other equation produces an equivalent system. This fact is used to solve systems of equations by the Elimination Method. Consider an example system of linear equations.

{3 x + 2 y = 6 y = 2 x - 11 (I) (II)

To solve the system by using the Elimination Method, there are five steps to follow.

Write the Equations in the Same Form

First, all like terms must be gathered on the same sides of the equations. In Equation (I), the variable terms are on the same side of the equation. However, the variable terms are on both sides of the equations in Equation (II). Like terms can be gathered on the same sides of the equations by applying the Properties of Equality.

{3 x + 2 y = 6 y = 2 x - 11

SubEqn

$(II):$ $LHS - 2 x = RHS - 2 x$

{3 x + 2 y = 6 - 2 x + y = - 11

Multiply an Equation

Multiply one of the equations by a constant so that one of the variable terms of the resulting equation is equal to or is the opposite of the corresponding variable term in the other equation. In this case, multiplying Equation (II) by

- 2

will produce opposite coefficients for the

y -

variable.

- 2 (- 2 x + y) = - 2 (- 11) ⇕ 4 x - 2 y = 22

Both the original and the resulting equations have the same solutions because they are equivalent equations.

Add the New Equation and the Other Equation

After the rewrites, the system of equations looks the following way.

{3 x + 2 y = 6 4 x - 2 y = 22

Add these two equations by adding the right-hand sides together and the left-hand sides together. This way one variable will be eliminated.

3 x + 2 y + (4 x - 2 y) = 6 + 22

▼

Simplify

RemovePar

Remove parentheses

3 x + 2 y + 4 x - 2 y = 6 + 22

CommutativePropAdd

Commutative Property of Addition

3 x + 4 x + 2 y - 2 y = 6 + 22

AddSubTerms

Add and subtract terms

7 x = 28

Note that this step results in an equation in only one variable. This equation can be solved by dividing both sides by

7 .

7 x = 28

DivEqn

$LHS / 7 = RHS / 7$

\frac{7 x}{7} = \frac{2 8}{7}

CalcQuot

Calculate quotient

x = 4

Write an Equivalent System

Substitute the value for the solved equation in one variable for any of the equations of the system. This produces an equivalent system of equations. In this case, Equation (I) will be replaced.

{3 x + 2 y = 6 - 2 x + y = - 11 (I) (II) ⇕ {x = 4 - 2 x + y = - 11

Note that the first equation is the solution value of

x .

Solve the Equivalent System

To solve the new system, substitute the found value into the other equation. In this case,

4

will be substituted into Equation (II) for

x

to find the value of

y .

{x = 4 - 2 x + y = - 11

Substitute

$(II):$ $x = 4$

{x = 4 - 2 (4) + y = - 11

▼

(II):

Solve for

y

MultNegPos

$(II):$ $(- a) b = - a b$

{x = 4 - 8 + y = - 11

AddEqn

$(II):$ $LHS + 8 = RHS + 8$

{x = 4 y = - 3

In this system, the value of

y

- 3 .

Therefore, the solution to the system of equations, which is the point of intersection of the lines, is

(4, - 3),

x = 4,

y = - 3 .

If at any step of the method a true statement is found, then the lines represented by the equations of the system are coincidental. The system would be a dependent system and have infinitely many solutions. Conversely, if at any step a false statement is found, then the lines are parallel. In this case, the system would be inconsistent and have no solution.

Be aware that when using this method, it is possible to make some mistakes in any of the steps. Therefore, it is important to check if the obtained solution is actually a solution to the original system.

Pop Quiz

Checking if the Coordinate Pair is a Solution

For each system of linear equations, verify whether the coordinate pair is a solution.

Random Check if Coordinate Pair is a Solution to a System of Equations

Example

The Price of Gold

A vlogger that Diego likes to watch bought silver and gold to make Olympic-style medals. The vlogger will show the process of making the medals as a multi-video series.

The price of gold is about

$ 2000

per ounce, while the price of of silver is about

$ 25

per ounce. The vlogger spent a total of

$ 7075

in both silver and gold. Let

g

and

s

be the ounces of gold and silver, respectively, that the vlogger bought. Then, the given information can be modeled by a linear equation.

25 s + 2000 g = 7075

Diego notes that if he subtracts

50

times the number of ounces of gold from

5

times the number of ounces of silver, there would still be

65

ounces left. This can also be written as an equation.

5 s - 50 g = 65

A system of equations is obtained by combining these two equations.

{25 s + 2000 g = 7075 5 s - 50 g = 65 (I) (II)

a Solve the system of equations graphically.

b Solve the system of equations using the Substitution Method.

c Solve the system of equations using the Elimination Method.

d Interpret the system in terms of consistency and independence.

Answer

a See solution.

b See solution.

c See solution.

d The system is consistent and independent.

Hint

a First write the equations in slope-intercept form.

b Isolate one variable in an equation.

c Recall that equivalent systems have the same solution.

d A system of equations can be inconsistent, consistent dependent, or consistent independent.

Solution

a To solve the system graphically, first each equation should be written in slope-intercept form. The variable

s

will arbitrarily be considered as the dependent variable, and

g

will be the independent variable.

{25 s + 2000 g = 7075 5 s - 50 g = 65 (I) (II)

▼

(I), (II):

Write in slope-intercept form

DivEqn

$(I):$ $LHS / 25 = RHS / 25$

{\frac{2 5 s + 2 0 0 0 g}{2 5} = 283 5 s - 50 g = 65

WriteSumFrac

$(I):$ Write as a sum of fractions

{\frac{2 5 s}{2 5} + \frac{2 0 0 0 g}{2 5} = 283 5 s - 50 g = 65

DivEqn

$(II):$ $LHS / 5 = RHS / 5$

{\frac{2 5 s}{2 5} + \frac{2 0 0 0 g}{2 5} = 283 \frac{5 s - 5 0 g}{5} = 13

WriteDiffFrac

$(II):$ Write as a difference of fractions

{\frac{2 5 s}{2 5} + \frac{2 0 0 0 g}{2 5} = 283 \frac{5 s}{5} - \frac{5 0 g}{5} = 13

$(I), (II):$ $\frac{a \cdot b}{c} = \frac{a}{c} \cdot b$

{\frac{2 5}{2 5} s + \frac{2 0 0 0}{2 5} g = 283 \frac{5}{5} s - \frac{5 0}{5} g = 13

$(I), (II):$ Calculate quotient

{1 s + 80 g = 283 1 s - 10 g = 13

$(I), (II):$ Identity Property of Multiplication

{s + 80 g = 283 s - 10 g = 13

SubEqn

$(I):$ $LHS - 80 g = RHS - 80 g$

{s = - 80 g + 283 s - 10 g = 13

AddEqn

$(II):$ $LHS + 10 g = RHS + 10 g$

{s = - 80 g + 283 s = 10 g + 13

Now, both equations can be graphed on the same coordinate plane.

Looking at the graph, it can be seen that the solution is about $3$ ounces of gold and a bit more than $40$ ounces of silver. In this case, since the point of intersection is not a lattice point, finding the exact solution is not possible.

b To solve the system using the Substitution Method, first a variable should be isolated from one of the equations. It seems easier to isolate

s

on Equation (II).

{25 s + 2000 g = 7075 5 s - 50 g = 65 (I) (II)

▼

(II):

Solve for

s

DivEqn

$(II):$ $LHS / 5 = RHS / 5$

{25 s + 2000 g = 7075 \frac{5 s - 5 0 g}{5} = 13

WriteDiffFrac

$(II):$ Write as a difference of fractions

{25 s + 2000 g = 7075 \frac{5 s}{5} - \frac{5 0 g}{5} = 13

MovePartNumRight

$(II):$ $\frac{a \cdot b}{c} = \frac{a}{c} \cdot b$

{25 s + 2000 g = 7075 \frac{5}{5} s - \frac{5 0}{5} g = 13

CalcQuot

$(II):$ Calculate quotient

{25 s + 2000 g = 7075 1 s - 10 g = 13

IdPropMult

$(II):$ Identity Property of Multiplication

{25 s + 2000 g = 7075 s - 10 g = 13

AddEqn

$(II):$ $LHS + 10 g = RHS + 10 g$

{25 s + 2000 g = 7075 s = 10 g + 13

Substitute

$(I):$ $s = 10 g + 13$

{25 (10 g + 13) + 2000 g = 7075 s = 10 g + 13

▼

(I):

Solve for

g

Distr

$(I):$ Distribute $25$

{250 g + 325 + 2000 g = 7075 s = 10 g + 13

AddTerms

$(I):$ Add terms

{2250 g + 325 = 7075 s = 10 g + 13

SubEqn

$(I):$ $LHS - 325 = RHS - 325$

{2250 g = 6750 s = 10 g + 13

DivEqn

$(I):$ $LHS / 2250 = RHS / 2250$

{g = 3 s = 10 g + 13

The vlogger bought

3

ounces of gold. It is possible to calculate the number of ounces of silver he purchased by substituting this value for

s

into Equation (II).

{g = 3 s = 10 g + 13 (I) (II)

Substitute

$(II):$ $g = 3$

{g = 3 s = 10 (3) + 13

▼

(II):

Solve for

s

Multiply

$(II):$ Multiply

{g = 3 s = 30 + 13

AddTerms

$(II):$ Add terms

{g = 3 s = 43

Therefore, the vlogger has

43

ounces of silver.

c To solve this system using the Elimination Method, it is important to remember that equivalent systems share the same solution. Equation (I) can be divided by

5

so that the variable

s

has the same coefficient in both equations.

{25 s + 2000 g = 7075 5 s - 50 g = 65 (I) (II)

▼

(I):

Simplify

DivEqn

$(I):$ $LHS / 5 = RHS / 5$

{\frac{2 5 s + 2 0 0 0 g}{5} = 1415 5 s - 50 g = 65

WriteSumFrac

$(I):$ Write as a sum of fractions

{\frac{2 5 s}{5} + \frac{2 0 0 0 g}{5} = 1415 5 s - 50 g = 65

MovePartNumRight

$(I):$ $\frac{a \cdot b}{c} = \frac{a}{c} \cdot b$

{\frac{2 5}{5} s + \frac{2 0 0 0}{5} g = 1415 5 s - 50 g = 65

CalcQuot

$(I):$ Calculate quotient

{5 s + 400 g = 1415 5 s - 50 g = 65

Now, to continue solving using the Elimination Method, Equation (II) will be subtracted from Equation (I).

{5 s + 400 g = 1415 5 s - 50 g = 65 (I) (II)

SysEqnSub

$(I):$ Subtract $(II)$

{5 s + 400 g - (5 s - 50 g) = 1415 - 65 5 s - 50 g = 65

▼

(I):

Solve for

g

Distr

$(I):$ Distribute $- 1$

{5 s + 400 g - 5 s + 50 g = 1415 - 65 5 s - 50 g = 65

AddSubTerms

$(I):$ Add and subtract terms

{450 g = 1350 5 s - 50 g = 65

DivEqn

$(I):$ $LHS / 450 = RHS / 450$

{g = 3 5 s - 50 g = 65

The vlogger has

3

ounces of gold this time, too. This value can be substituted into Equation (II) to find out how many ounces of silver he has.

{g = 3 5 s - 50 g = 65 (I) (II)

Substitute

$(II):$ $g = 3$

{g = 3 5 s - 50 (3) = 65

▼

(II):

Solve for

s

Multiply

$(II):$ Multiply

{g = 3 5 s - 150 = 65

AddEqn

$(II):$ $LHS + 150 = RHS + 150$

{g = 3 5 s = 215

DivEqn

$(II):$ $LHS / 5 = RHS / 5$

{g = 3 s = 43

In this case, using the Elimination Method might be a little more intuitive than using the Substitution Method. Note that both of these methods are more reliable than solving the system graphically.

d To interpret the system in terms of the independence and consistency, these concepts should be reviewed.

Concept	Definition
Consistent System	A system of equations that has at least one solution.
Inconsistent System	A system of equations that has no solution.
Dependent System	A system of equations with infinitely many solutions.
Independent System	A system of equations that has exactly one solution.

Since the system has exactly one solution, it is both consistent and independent.

Example

Buying Equipment for the School Gym

As part of his student council duties, Tadeo is in charge of getting equipment for the school gym. At the moment, the gym needs new basketballs and volleyballs. Each basketball costs $$ 75$ and each volleyball costs $$ 50 .$ The gym's budget is $$ 1400 .$

Tadeo can only go to the store once to buy the equipment. Each basketball weighs

22

ounces and each volleyball weighs

10

ounces. The total weight of the purchased balls is

364

ounces. For Tadeo to be able to spend all the budget in one go, the number of basketballs

b

and volleyballs

v

need to satisfy the following system of linear equations.

{75 b + 50 v = 1400 22 b + 10 v = 364 (I) (II)

a Solve the system using the Elimination Method.

b Verify the answer from Part A.

c Interpret the system in terms of the consistency and independence.

Answer

a See solution.

b See solution.

c The system is consistent and independent.

Hint

a What is the easiest way for a variable to have the same coefficient in both equations?

b Substitute the values from Part A into the original system of equations.

c A system of equations can be inconsistent, consistent independent, or consistent dependent.

Solution

a To solve the system using the Elimination Method, the coefficients of one variable have to be same to eliminate the term by adding or subtracting. Looking at the system, it can be noted that dividing Equation (I) by

- 5

will achieve this purpose.

{75 b + 50 v = 1400 22 b + 10 v = 364 (I) (II)

▼

(I):

Simplify

DivEqn

$(I):$ $LHS / (- 5) = RHS / (- 5)$

{\frac{7 5 b + 5 0 v}{- 5} = - 280 22 b + 10 v = 364

WriteSumFrac

$(I):$ Write as a sum of fractions

{\frac{7 5 b}{- 5} + \frac{5 0 v}{- 5} = - 280 22 b + 10 v = 364

MovePartNumRight

$(I):$ $\frac{a \cdot b}{c} = \frac{a}{c} \cdot b$

{\frac{7 5}{- 5} b + \frac{5 0}{- 5} v = - 280 22 b + 10 v = 364

MoveNegDenomToFrac

$(I):$ Put minus sign in front of fraction

{- \frac{7 5}{5} b + (- \frac{5 0}{5} v) = - 280 22 b + 10 v = 364

AddNeg

$a + (- b) = a - b$

{- \frac{7 5}{5} b - \frac{5 0}{5} v = - 280 22 b + 10 v = 364

CalcQuot

$(I):$ Calculate quotient

{- 15 b - 10 v = - 280 22 b + 10 v = 364

Now,

v

has opposite coefficients. Therefore, by adding the equations, this variable will be eliminated.

{- 15 b - 10 v = - 280 22 b + 10 v = 364 (I) (II)

SysEqnAdd

$(I):$ Add $(II)$

{- 15 b - 10 v + (22 b + 10 v) = - 280 + 364 22 b + 10 v = 364

▼

(I):

Solve for

b

RemovePar

$(I):$ Remove parentheses

{- 15 b - 10 v + 22 b + 10 v = - 280 + 364 22 b + 10 v = 364

AddSubTerms

$(I):$ Add and subtract terms

{7 b = 84 22 b + 10 v = 364

DivEqn

$(I):$ $LHS / 7 = RHS / 7$

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Catch-Up and Review

Proof

Answer

Hint

Solution

Answer

Hint

Solution