Solve Systems of Equations Using Elimination Method

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

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

$(I):$ 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

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

{\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

$(I), (II):$ \IdPropMult

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

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

{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

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

{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

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

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

$(II):$ Calculate quotient

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

IdPropMult

$(II):$ \IdPropMult

{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

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

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

(I):

Solve for

g

$(I):$ Distribute $25$

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

AddTerms

$(I):$ Add terms

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

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

{2250 g = 6750 s = 10 g + 13

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

{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)

$(II):$ $g = 3$

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

(II):

Solve for

s

$(II):$ 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

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

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

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

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

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

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

$(I):$ 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)

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

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

(I):

Solve for

g

$(I):$ 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

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

{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)

$(II):$ $g = 3$

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

(II):

Solve for

s

$(II):$ Multiply

{g = 3 5 s - 150 = 65

AddEqn

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

{g = 3 5 s = 215

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

{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

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

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

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

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

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

{\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

$(I):$ 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

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

{b = 12 22 b + 10 v = 364

Since

b = 12,

Tadeo can buy and carry

12

basketballs. To find the number of volleyballs he can buy, this value will be substituted into Equation (II).

{b = 12 22 b + 10 v = 364 (I) (II)

$(II):$ $b = 12$

{b = 12 22 (12) + 10 v = 364

(II):

Solve for

v

$(II):$ Multiply

{b = 12 264 + 10 v = 364

$(II):$ $LHS - 264 = RHS - 264$

{b = 12 10 v = 100

$(II):$ $LHS / 10 = RHS / 10$

{b = 12 v = 10

Therefore, Tadeo can buy and carry

12

basketballs and

10

volleyballs.

b To verify the solution, the values found in Part A need to be substituted into the system of equations.

	Equation (I)	Equation (II)
Equation	$75 b + 50 v = 1400$	$22 b + 10 v = 364$
Substitute	$75 (12) + 50 (10) = 1400$	$22 (12) + 10 (10) = 364$
Simplify	$1400 = 1400 ✓$	$364 = 364 ✓$

The values verify both equations of the system. Therefore, the solution is correct.

c To interpret the system in terms of independence and consistency, these concepts will 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, the system is both consistent and independent.

Example

Tools in a Toolbox

Ignacio is looking through his dad's toolbox to see if there are any duplicates that can be sold at a garage sale. He discovers that there are some extra hammers and wrenches.

In fact, the number of wrenches is equal to the difference between

13

and twice the number of hammers. He figures out that if he sells each hammer at

$ 7

and each wrench at

$ 3.50,

he will make a profit of

$ 45.50 .

h

and

w

are the number of hammers and wrenches, respectively, the information can be written as a system of equations.

{w = 13 - 2 h 7 h + 3.50 w = 45.50 (I) (II)

a Solve the system of equations using the Elimination Method.

b Interpret the system in terms of consistency and independence.

Answer

a See solution.

b The system is consistent and dependent.

Hint

a In Equation (I), group the variable terms on the same side.

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

Solution

a To solve the system using the Elimination Method, all the variables should be on the same side of the equation in both equations. This can be achieved by adding

2 h

to both sides of Equation (I).

{w = 13 - 2 h 7 h + 3.50 w = 45.50 (I) (II) ⇕ {2 h + w = 13 7 h + 3.50 w = 45.50 (I) (II)

Now, a variable needs to have the same or opposite coefficients in both equations. This can be achieved by multiplying Equation (I) by

3.5 .

{2 h + w = 13 7 h + 3.5 w = 45.5 (I) (II)

MultEqn

$(I):$ $LHS \cdot 3.5 = RHS \cdot 3.5$

{(2 h + w) 3.5 = 45.5 7 h + 3.5 w = 45.5

$(I):$ Distribute $3.5$

{7 h + 3.5 w = 45.5 7 h + 3.5 w = 45.5

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

{7 h + 3.5 w = 45.5 7 h + 3.5 w - (7 h + 3.5 w) = 45.5 - 45.5

$(II):$ Distribute $- 1$

{7 h + 3.5 w = 45.5 7 h + 3.5 w - 7 h - 3.5 w = 45.5 - 45.5

SubTerms

$(II):$ Subtract terms

{7 h + 3.5 w = 45.5 0 = 0 ✓

By doing these operations, Equation (II) becomes an identity, or a true statement. This means that this system of equations has infinitely many solutions. Realizing this, Ignacio thinks that it is too big of a coincidence and decides to count the tools again and change the prices before going to the garage sale.

b To interpret the system in terms of consistency and independence, 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 infinitely many solutions, the system is consistent and dependent. In a system of linear equations, if the system is dependent, then the equations that make the system are equations of coincidental lines.

Example

The Lengths of Two Tracks

LaShay is on the track and field team at school. She has two tracks nearby her home where she trains for races. She is trying to remember the lengths of the tracks.

If LaShay completes

2

laps on the first track and

1

lap on the second track, then she runs

5

miles. On the other hand, if she completes

6

laps on the first track and

3

laps on the second track, then she runs

18

miles. Letting

x

be the length of the first track and

y

the length of the second track, this information can be written as a system of linear equations.

{2 x + y = 5 6 x + 3 y = 18 (I) (II)

a Solve the system of equations using the Elimination Method.

b Interpret the system in terms of consistency and independence.

Answer

a See solution.

b The system is inconsistent.

Hint

a Make the coefficients of a variable term the same in both equations.

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

Solution

a To solve the system using the Elimination Method, a variable term first needs to have the same coefficient in both equations. To achieve this, the first equation will be multiplied by

3 .

{2 x + y = 5 6 x + 3 y = 18 (I) (II)

(I):

Simplify

MultEqn

$(I):$ $LHS \cdot 3 = RHS \cdot 3$

{(2 x + y) 3 = 15 6 x + 3 y = 18

$(I):$ Distribute $3$

{6 x + 3 y = 15 6 x + 3 y = 18

The

y -

variable has the same coefficient in both equations. Now, Equation (II) can be subtracted from Equation (I).

{6 x + 3 y = 15 6 x + 3 y = 18 (I) (II)

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

{6 x + 3 y - (6 x + 3 y) = 15 - 18 6 x + 3 y = 18

(I):

Simplify

$(I):$ Distribute $- 1$

{6 x + 3 y - 6 x - 3 y = 15 - 18 6 x + 3 y = 18

SubTerms

$(I):$ Subtract terms

{0 = - 3 \times 6 x + 3 y = 18

Equation (I) was reduced to a false statement. Because of this, the equation has no solution. Knowing this, LaShay realizes that she does not remember the lengths of the tracks and decides to properly measure the tracks to prepare her training plan.

b To interpret the system in terms of consistency and independence, 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 no solution, it is an inconsistent system.

Pop Quiz

Practice Solving Systems of Linear Equations

Solve the system of linear equations to find the values of $x$ and $y .$

Closure

Finding Out How Many Toys Are in Each Box

At the beginning of this lesson, a system of equations was introduced for the number of toys in a box. In both equations,

x

is the number of robots and

y

the number of cars.

{\frac{4}{5} x + \frac{1}{1 0} y = 20 15 x + 5 y = 500 (I) (II)

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

Hint

Multiply or divide an equation by a number so that a variable has the same coefficient in both equations.

Solution

This system of linear equations can be solved without graphing using the Elimination Method. To use this method, Equation (I) will be multiplied by

10

to get rid of the fractions, while Equation (II) will be divided by

5

to make the coefficients of the

y -

variable equal.

{\frac{4}{5} x + \frac{1}{1 0} y = 20 15 x + 5 y = 500 (I) (II)

(I), (II):

Simplify

MultEqn

$(I):$ $LHS \cdot 10 = RHS \cdot 10$

{(\frac{4}{5} x + \frac{1}{1 0} y) 10 = 200 15 x + 5 y = 500

$(I):$ Distribute $10$

{\frac{4}{5} x (10) + \frac{1}{1 0} y (10) = 200 15 x + 5 y = 500

CommutativePropMult

$(I):$ \CommutativePropMult

{\frac{4}{5} (10) x + \frac{1}{1 0} (10) y = 200 15 x + 5 y = 500

MoveRightFacToNum

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

{\frac{4 0}{5} x + \frac{1 0}{1 0} y = 200 15 x + 5 y = 500

$(I):$ Calculate quotient

{8 x + 1 y = 200 15 x + 5 y = 500

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

{8 x + 1 y = 200 \frac{1 5 x + 5 y}{5} = 100

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

{8 x + 1 y = 200 \frac{1 5 x}{5} + \frac{5 y}{5} = 100

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

{8 x + 1 y = 200 \frac{1 5}{5} x + \frac{5}{5} y = 100

$(II):$ Calculate quotient

{8 x + 1 y = 200 3 x + 1 y = 100

$(I), (II):$ \IdPropMult

{8 x + y = 200 3 x + y = 100

Now that the variable

y

has the same coefficient in both equations, Equation (II) will be subtracted from Equation (I). Then, Equation (I) will be solved for the

x -

variable.

{8 x + y = 200 3 x + y = 100 (I) (II)

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

{8 x + y - (3 x + y) = 200 - 100 3 x + y = 100

(I):

Solve for

x

$(I):$ Distribute $- 1$

{8 x + y - 3 x - y = 200 - 100 3 x + y = 100

SubTerms

$(I):$ Subtract terms

{5 x = 100 3 x + y = 100

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

{x = 20 3 x + y = 100

Therefore, there are

20

robots in each box. Now, the value of

x

will be substituted into Equation (II) to find the value of

y .

{x = 20 3 x + y = 100

$(II):$ $x = 20$

{x = 20 3 (20) + y = 100

(II):

Solve for

y

$(II):$ Multiply

{x = 20 60 + y = 100