Algebraic Solutions to Systems of Equations: Substitution and Elimination

Systems of equations are used to relate the values of two or more variables. There are different methods of solving a system of equations. This lesson will present these methods and show how to use them.

Catch-Up and Review

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

Challenge

Books and Movies About Space

Vincenzo is fascinated by all things related to space and astronauts. He spends a lot of his free time reading books and watching movies about space travel, distant galaxies, and rocket science.

Vincenzo counted that he has watched or read

27

things related to space movies or books in total. The number of movies he has seen is

9

more than the number of books he has read. What are the numbers of movies and books about space that Vincenzo had watched or read?

Discussion

Substitution Method of Solving a System of Equations

There are several methods for solving a system of equations. One of the most popular methods is the Substitution Method.

Method

Substitution Method

The Substitution Method is an algebraic method for finding the solutions of a system of equations. It consists of substituting an equivalent expression for a variable in one of the equations of the system. Consider, for example, the following system of linear equations.

{y - 4 = 2 x 9 x + 6 = 3 y (I) (II)

To solve the system by using the Substitution Method, there are four steps to follow.

Isolate One Variable in Any of the Equations

The first step is to isolate any variable in any of the equations. For simplicity, in this case, the

y -

variable will be isolated in Equation (I).

{y - 4 = 2 x 9 x + 6 = 3 y

AddEqn

$(I):$ $LHS + 4 = RHS + 4$

{y = 2 x + 4 9 x + 6 = 3 y

Substitute the Expression

Substitute the new expression for the variable in the equation where the variable was not isolated. In this case,

2 x + 4

will be substituted for

y

in Equation (II).

{y = 2 x + 4 9 x + 6 = 3 y

Substitute

$(II):$ $y = 2 x + 4$

{y = 2 x + 4 9 x + 6 = 3 (2 x + 4)

Now Equation (II) only has one variable, which is

x .

Solve the Equation With One Variable

Solve the equation that contains only one variable. In this case, Equation (II) will be solved for

x .

{y = 2 x + 4 9 x + 6 = 3 (2 x + 4) (I) (II)

Distr

$(II):$ Distribute $3$

{y = 2 x + 4 9 x + 6 = 6 x + 12

SubEqn

$(II):$ $LHS - 6 = RHS - 6$

{y = 2 x + 4 9 x = 6 x + 6

SubEqn

$(II):$ $LHS - 6 x = RHS - 6 x$

{y = 2 x + 4 3 x = 6

DivEqn

$(II):$ $LHS / 3 = RHS / 3$

{y = 2 x + 4 x = 2

The value of the

x -

variable is

2 .

Substitute the Value of the Variable Into the Other Equation

Now that the value of one of the variables is known, it can be substituted into the equation that has not been considered yet. Here,

x = 2

will be substituted into Equation (I).

{y = 2 x + 4 x = 2 (I) (II)

Substitute

$(I):$ $x = 2$

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

▼

(I):

Evaluate right-hand side

Multiply

$(I):$ Multiply

{y = 4 + 4 x = 2

AddTerms

$(I):$ Add terms

{y = 8 x = 2

The value of the

y -

variable in this system is

8 .

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

(2, 8)

x = 2,

y = 8 .

Example

First Time Spacewalking

After reading another book about space, Vincenzo quickly fell asleep and dreamed that he was an astronaut spacewalking for the first time. What an amazing experience!

External credits: @catalyststuff

Vincenzo received a task to install some external parts to the spaceship. The number of parts

p

he installed and the number of minutes

m

he spent in the open space are related by a system of equations.

{p + 4 m = 42 8 p - 5 m = 3

a Solve the system by graphing.

b Solve the system by substitution.

c Are the solutions the same? Which method of solving is more useful in this case and why?

Answer

a Graph:

Solution: $m = 9,$ $p = 6$

m = 9,

p = 6

c The solutions are the same. In this case, the Substitution Method is more useful for a number of reasons.

Hint

a Rewrite the equations in slope-intercept form. Then use the slope and

y -

intercept to graph each equation.

b Isolate

p

in the first equation and substitute the corresponding expression into the second equation to find

m .

Then substitute the value of

m

into the first equation and find

p .

c Identify which method is shorter. Does either method require the equations to be in a specific form? Do they both result in finding the exact solutions every time?

Solution

a In order to solve the system of equations by graphing, both equations should be written in slope-intercept form.

y = m x + b

Rewrite both equations until they match this form. Notice that the first equation is already almost in this form — all that is left is to subtract

4 m

from both sides.

p + 4 m - 4 m = 42 - 4 m ⇕ p = - 4 m + 42

Rewrite the second equation similarly.

8 p - 5 m = 3

▼

Write in slope-intercept form

AddEqn

$LHS + 5 m = RHS + 5 m$

8 p = 3 + 5 m

DivEqn

$LHS / 8 = RHS / 8$

p = \frac{3 + 5 m}{8}

WriteSumFrac

Write as a sum of fractions

p = \frac{3}{8} + \frac{5 m}{8}

MovePartNumRight

$\frac{a \cdot b}{c} = \frac{a}{c} \cdot b$

p = \frac{3}{8} + \frac{5}{8} m

CommutativePropAdd

Commutative Property of Addition

p = \frac{5}{8} m + \frac{3}{8}

Now, graph both equations on the same coordinate plane. To graph the first equation, start by plotting the

y -

intercept of

42 .

Next, use the slope of

- 4

to move

1

unit to the right and

4

units down, or

2

units to the right and

2 \cdot 4 = 8

units down, to plot the second point.

Draw a line through the two plotted points to get the graph of the first equation.

The second equation can be graphed by following the same process.

The solution of the system of equations is represented by the point of intersection of the lines. If the point of intersection lies on lattice lines or their intersections, the exact solution will be determined. Otherwise, only an estimate of the solution might be found.

The lines intersect at $(9, 6) .$ Therefore, $m = 9$ and $p = 6,$ which indicates that Vincenzo spent $9$ minutes spacewalking and installed $6$ parts on the spaceship.

b The system of equations will now be solved by using the Substitution Method. Start by isolating the variable

p

and substituting the corresponding expression into the second equation.

{p + 4 m = 42 8 p - 5 m = 3 (I) (II)

SubEqn

$(I):$ $LHS - 4 m = RHS - 4 m$

{p = 42 - 4 m 8 p - 5 m = 3

Substitute

$(II):$ $p = 42 - 4 m$

{p = 42 - 4 m 8 (42 - 4 m) - 5 m = 3

Distr

$(II):$ Distribute $8$

{p = 42 - 4 m 336 - 32 m - 5 m = 3

SubTerm

$(II):$ Subtract term

{p = 42 - 4 m 336 - 37 m = 3

SubEqn

$(II):$ $LHS - 336 = RHS - 336$

{p = 42 - 4 m - 37 m = - 333

DivEqn

$(II):$ $LHS / (- 37) = RHS / (- 37)$

{p = 42 - 4 m m = 9

The value of

m

is found to be

9 .

Now it can be substituted in either of the original equations. Notice that

p

is already isolated in the first equation, so it might be convenient to substitute the value of

m

into this equation and evaluate

p .

{p = 42 - 4 m m = 9

Substitute

$(I):$ $m = 9$

{p = 42 - 4 (9) m = 9

Multiply

$(I):$ Multiply

{p = 42 - 36 m = 9

SubTerm

$(I):$ Subtract term

{p = 6 m = 9

The solution to the system of equations is

p = 6

and

m = 9 .

c Both methods of solving the system of equations gave the same solution. Therefore, both methods of solving are correct.

Graphing Method ↘ (9, 6) Substitution Method ↙

However, in this case, the Substitution Method can be more convenient because it is shorter and gives the exact solution. By comparison, the graphing method requires the equations to be in slope-intercept form and does not always result in finding the exact solution.

Example

Distances Between Planets

In his dreams, Vincenzo gets to travel to planets far far away. Traveling to two distant planets Lunaris and Exosia from Earth takes

137

years and

680

years, respectively.

The distances from Earth to Lunaris

ℓ

and to Exosia

e

are given by the following system of equations.

{2 ℓ + 2 = e ℓ + 3 e = 97

a Solve the system by using the Substitution Method.

b Check the solution by substituting it into both equations of the system.

c Graph the system of equations and analyze the coordinates of the point of intersection.

Answer

ℓ = 13

and

e = 28

b See solution.

c Graph:

Hint

a Isolate one variable in one of the equations. Substitute the corresponding expression into the other equation to solve for the other variable.

b Substitute the solution from Part A into the system of equations and see if true statements are found.

c Rewrite each equation in slope-intercept form. Then, graph the equations using the

y -

intercepts and slopes.

Solution

a The system of equations will be solved by using the Substitution Method. Start by isolating one variable in one equation. Notice that

e

is already isolated in the first equation.

{2 ℓ + 2 = e ℓ + 3 e = 97 ⇕ {e = 2 ℓ + 2 ℓ + 3 e = 97

Substitute the corresponding expression into the other equation. Then, solve for the other variable.

{e = 2 ℓ + 2 ℓ + 3 e = 97 (I) (II)

Substitute

$(II):$ $e = 2 ℓ + 2$

{e = 2 ℓ + 2 ℓ + 3 (2 ℓ + 2) = 97

▼

Solve for

ℓ

Distr

$(II):$ Distribute $3$

{e = 2 ℓ + 2 ℓ + 6 ℓ + 6 = 97

AddTerms

$(II):$ Add terms

{e = 2 ℓ + 2 7 ℓ + 6 = 97

SubEqn

$(II):$ $LHS - 6 = RHS - 6$

{e = 2 ℓ + 2 7 ℓ = 91

DivEqn

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

{e = 2 ℓ + 2 ℓ = 13

It was calculated that

ℓ

equals

13 .

Next, substitute this value into either of the original equations and solve for the other variable

e .

In this case, the first equation will be used since

e

is already isolated on one side.

{e = 2 ℓ + 2 ℓ = 13

Substitute

$(I):$ $ℓ = 13$

{e = 2 (13) + 2 ℓ = 13

Multiply

$(I):$ Multiply

{e = 26 + 2 ℓ = 13

SubTerm

$(I):$ Subtract term

{e = 28 ℓ = 13

The solution to the system is

ℓ = 13

and

e = 28 .

b To check the solution, substitute

13

for

ℓ

and

28

for

e

into the system of equations. If both equations result in true statements after simplification, the solution is correct.

2 ℓ + 2 = e ℓ + 3 e = 97

SubstituteII

$ℓ = 13$ , $e = 28$

2 (13) + 2 = ? 28 13 + 3 (28) = ? 97

Multiply

26 + 2 = ? 28 13 + 84 = ? 97

AddTerms

Add terms

28 = 28 ✓ 97 = 97 ✓

The equations both simplified into true statements, so the solution is indeed correct!

c To solve the system of equations by graphing, start by rewriting both equations in slope-intercept form. Consider

ℓ

as the

y -

variable and

e

as the

x -

variable. Start with Equation (I).

2 ℓ + 2 = e

SubEqn

$LHS - 2 = RHS - 2$

2 ℓ = e - 2

DivEqn

$LHS / 2 = RHS / 2$

ℓ = 0.5 e - 1

Similarly, rewrite Equation (II) in slope-intercept form.

ℓ + 3 e = 97 ⇓ ℓ = - 3 e + 97

Now, graph the equations using their

y -

intercepts and slopes. The point of intersection represents the solution.

The point of intersection lies on a lattice line where $e = 28 .$ However, it can be difficult to determine the exact value of $ℓ$ just by looking at the graph. It can have values from $11$ to $14 .$ In Part A it was found that $ℓ$ is $13 .$ The graph does support that value, so the solution is $(28, 13) .$

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 .

Example

Pit Stop on Exosia

Vincenzo and his team reached the planet Exosia and made a short stop there to refuel and repair their spaceship. The people of Exosia help Vincenzo and his crew make some modifications to their ship so they can travel at even greater speeds!

Astronaut with a spaceship on the planet Exosia

External credits: @catalyststuff

Their initial maximum speed

s

and the improved intergalactic superspeed

n

are related by the following system of equations.

{2 n - s = 37 7 s - 4 n = - 19

a Solve this system by graphing.

b Solve the same system by elimination.

c Are the solutions the same?

Answer

a Graph:

s = 11,

n = 24

c Yes

Hint

a Rewrite each equation in slope-intercept form. Then, graph both equations on the same coordinate plane.

b Multiply the first equation by

2

and add the equations together to eliminate

n

and solve for

s .

c Compare the solutions found by each method.

Solution

a The system of equations can be solved graphically by first rewriting each equation in slope-intercept form. Consider

n

as the

y -

variable and

s

as the

x -

variable.

{2 n - s = 37 7 s - 4 n = - 19 (I) (II)

▼

(I), (II):

Write in slope-intercept form

AddEqn

$(I):$ $LHS + s = RHS + s$

{2 n = s + 37 7 s - 4 n = - 19

DivEqn

$(I):$ $LHS / 2 = RHS / 2$

{n = 0.5 s + 18.5 7 s - 4 n = - 19

SubEqn

$(II):$ $LHS - 7 s = RHS - 7 s$

{n = 0.5 s + 18.5 - 4 n = - 7 s - 19

DivEqn

$(II):$ $LHS / (- 4) = RHS / (- 4)$

{n = 0.5 s + 18.5 n = 1.75 s + 4.75

Both equations can now be graphed on the same coordinate plane.

Looking at the graph, the solution appears to be $s = 11$ and $n = 24 .$

b Notice that Equation (I) can be multiplied by

2

so that the

n -

terms can be eliminated by adding the equations.

{2 n - s = 37 7 s - 4 n = - 19 (I) (II)

MultEqn

$(I):$ $LHS \cdot 2 = RHS \cdot 2$

{4 n - 2 s = 74 7 s - 4 n = - 19

SysEqnAdd

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

{4 n - 2 s + (7 s - 4 n) = 74 + (- 19) 7 s - 4 n = - 19

▼

(I):

Solve for

s

AddNeg

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

{4 n - 2 s + 7 s - 4 n = 74 - 19 7 s - 4 n = - 19

AddSubTerms

$(I):$ Add and subtract terms

{5 s = 55 7 s - 4 n = - 19

DivEqn

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

{s = 11 7 s - 4 n = - 19

Equation (I) simplified to

s = 11 .

This value can be substituted into Equation (II) to calculate the value of

n .

{s = 11 7 s - 4 n = - 19 (I) (II)

Substitute

$(II):$ $s = 11$

{s = 11 7 (11) - 4 n = - 19

▼

(II):

Solve for

n

Multiply

$(II):$ Multiply

{s = 11 77 - 4 n = - 19

SubEqn

$(II):$ $LHS - 77 = RHS - 77$

{s = 11 - 4 n = - 96

DivEqn

$(II):$ $LHS / (- 4) = RHS / (- 4)$

{s = 11 n = 24

The solution is

s = 11

and

n = 24 .

Keep in mind that the Elimination Method works because equivalent systems share the same solution.

c Finally, the solutions found by using the two different methods can be compared.

Graphing Method : Elimination Method : s = 11, n = 24 s = 11, n = 24

Both methods resulted in the same solution, which means that they are both correct. Comparing the methods, using the Elimination Method might be a little easier and quicker than graphing the equations. This method also always results in finding the exact solution, while graphing sometimes results in finding just an estimation of the solution.

Example

Discovering a New Galaxy

After refueling and repairing the spaceship, Vincenzo continued his way across space. His destination is a new galaxy called the Stellar Nebula.

External credits: @pikisuperstar, @catalyststuff

Vincenzo used a laser measuring device on the spaceship to determine the dimensions of the galaxy. Its width

w

and height

h

are related by the following system of equations.

{3 w - 4 h = 6 5 h = 78 - w

a Solve the system by using the Elimination Method.

b Check the solution by substituting it in both equations of the system.

Answer

w = 18,

h = 12

c See solution.

Hint

a Add

w

to both sides of the second equation, then multiply it by

3 .

Subtract the equations to eliminate

w .

Solve the resulting equation for

h .

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

Solution

a Start by recalling that the Elimination Method is a method of eliminating one variable from a system of equations. This is done by first rewriting the coefficients to be the same or opposites, then adding or subtracting the equations. First, rewrite Equation (II) by adding

w

to both sides and multiplying by

3 .

{3 w - 4 h = 6 5 h = 78 - w (I) (II)

AddEqn

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

{3 w - 4 h = 6 5 h + w = 78

MultEqn

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

{3 w - 4 h = 6 15 h + 3 w = 234

SysEqnSub

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

{3 w - 4 h = 6 15 h + 3 w - (3 w - 4 h) = 234 - 6

▼

(I):

Solve for

h

Distr

$(II):$ Distribute $- 1$

{3 w - 4 h = 6 15 h + 3 w - 3 w - (- 4 h) = 234 - 6

SubNeg

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

{3 w - 4 h = 6 15 h + 3 w - 3 w + 4 h = 234 - 6

CommutativePropAdd

$(II):$ Commutative Property of Addition

{3 w - 4 h = 6 15 h + 4 h + 3 w - 3 w = 234 - 6

AddSubTerms

$(II):$ Add and subtract terms

{3 w - 4 h = 6 19 h = 228

DivEqn

$(II):$ $LHS / 19 = RHS / 19$

{3 w - 4 h = 6 h = 12

The value

h = 12

can be substituted into Equation (I) to calculate the value of

w .

{3 w - 4 h = 6 h = 12

Substitute

$(I):$ $h = 12$

{3 w - 4 (12) = 6 h = 12

▼

(I):

Solve for

w

Multiply

$(I):$ Multiply

{3 w - 48 = 6 h = 12

AddEqn

$(I):$ $LHS + 48 = RHS + 48$

{3 w = 54 h = 12

DivEqn

$(I):$ $LHS / 3 = RHS / 3$

{w = 18 h = 12

The solution is

w = 18

and

h = 12 .

This means that the galaxy is

18

galactic units wide and

12

galactic units high.

b To verify the solution, substitute the values found in Part A into the system of equations and simplify.

	Equation (I)	Equation (II)
Equation	$3 w - 4 h = 6$	$5 h = 78 - w$
Substitute	$3 (18) - 4 (12) = ? 6$	$5 (12) = ? 78 - 18$
Simplify	$6 = 6 ✓$	$60 = 60 ✓$

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

Pop Quiz

Checking the Solutions of Systems of Equations

Consider the given system of linear equations. Check whether the values of $x$ and $y$ correspond to a solution to the system.

System of equations and its solutions are randomly generated

Pop Quiz

Practice Solving Systems of Linear Equations

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

Discussion

Numbers of Solution to Systems of Equations

Solving a system of equations can result in three different scenarios. One possible scenario is when a system of equations has exactly one solution.

{y = 4 x + 5 7 x - y = 4 ⇓ x = 3 y = 17

The graph of this system of equations consists of two intersecting lines. The coordinates of the intersection point correspond to the solution of the system of equations.

Another possible scenario is when solving a system of equations results in an identity.

{y = 4 x + 5 2 y - 8 x = 10 \Leftrightarrow {y = 4 x + 5 10 = 10 ✓

In this case, the system of equations has infinitely many solutions and the graph of the system is two coincidental lines.

The last possible scenario is when solving a system of equations results in a false statement.

{y = 4 x + 5 2 x - 0.5 y = 3 \Leftrightarrow {y = 4 x + 5 - 2.5 = 3 \times

This means that the system of equations has no solution. The graph of this type of system of equations is two parallel lines.

These three scenarios are summarized in a table.

Number of Solutions	Graph
One solution	Intersecting lines
Infinitely many solutions	Coincidental lines
No solution	Parallel lines

Example

Stars and Planets in the Stellar Nebula

Vincenzo was amazed by the beauty of the newly discovered galaxy Stellar Nebula. It shined with blue and purple colors as he approached in his spaceship.

The Stellar Nebula has

p

million planets and

s

million stars. These numbers are related by the following system of equations.

{6 s - 15 p = 57 19 + 5 p = 2 s

a Solve the system of equations.

b Graph the system of equations on a coordinate plane. How do the lines relate to each other?

Answer

a Infinitely many solutions

b The lines are coincidental.

Hint

a Solve the system of equations by using the Substitution or Elimination Methods.

b Rewrite the equations in slope-intercept form and graph them. How many common points do their graphs have?

Solution

a Start by analyzing the given system of equations.

{6 s - 15 p = 57 19 + 5 p = 2 s (I) (II)

Notice that multiplying Equation (II) by

3,

rearranging the sides of that equation, and then subtracting it from Equation (I) will eliminate the variable

s .

This means that the Elimination Method can be used to solve the system of equations.

{6 s - 15 p = 57 19 + 5 p = 2 s (I) (II)

MultEqn

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

{6 s - 15 p = 57 57 + 15 p = 6 s

RearrangeEqn

$(II):$ Rearrange equation

{6 s - 15 p = 57 6 s = 57 + 15 p

SysEqnSub

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

{6 s - 15 p - 6 s = 57 - (57 + 15 p) 6 s = 57 + 15 p

SubTerm

$(I):$ Subtract term

{- 15 p = - 15 p 6 s = 57 + 15 p

An equation with the same expression on both sides was obtained, which means that it is a true statement for any value of the variable

p .

This means that the system of equations has infinitely many solutions.

b To graph the system of equations on a coordinate plane, first, rewrite both equations so that they are in slope-intercept form. Consider

s

as the

y -

variable and

p

as the

x -

variable.

{6 s - 15 p = 57 19 + 5 p = 2 s (I) (II)

▼

(I), (II):

Write in slope-intercept form

AddEqn

$(I):$ $LHS + 15 p = RHS + 15 p$

{6 s = 15 p + 57 19 + 5 p = 2 s

DivEqn

$(I):$ $LHS / 6 = RHS / 6$

{s = 2.5 p + 9.5 19 + 5 p = 2 s

RearrangeEqn

$(II):$ Rearrange equation

{s = 2.5 p + 9.5 2 s = 19 + 5 p

DivEqn

$(II):$ $LHS / 2 = RHS / 2$

{s = 2.5 p + 9.5 s = 9.5 + 2.5 p

CommutativePropAdd

$(II):$ Commutative Property of Addition

{s = 2.5 p + 9.5 s = 2.5 p + 9.5

Rewriting the equations resulted in the exact same equation. Graph it using the

y -

intercept of

9.5

and the slope of

2.5 .

Since the lines have the same equation, their graphs are coincidental lines. This piece of information highlights the fact that the lines have infinitely many common points. This means the system of equations has infinitely many solutions.

Example

Exploring a Black Hole

While exploring the new galaxy, Vincenzo and his team noticed a black hole on the edge of the galaxy. Curious, they flew closer to the black hole to register some of its characteristics.

A spaceship is approaching a black hole.

External credits: Ute Kraus, @catalyststuff

They tried to measure the density

d

and mass

m

of the black hole and got the following system of equations.

{16 d = - 8 m + 20 m = 4 - 2 d

a Solve the system of equations.

b Graph the system of equations on a coordinate plane. How do the lines relate to each other?

Answer

a No solution

b The lines are parallel.

Hint

a Solve the system of equations by using the Substitution or the Elimination Method.

b Rewrite the equations in slope-intercept form and graph them. How many common points do their graphs have?

Solution

a Start by analyzing the given system of equations.

{16 d = - 8 m + 20 m = 4 - 2 d (I) (II)

Note that in Equation (II), the variable

m

is isolated on the left-hand side. Substitute the corresponding expression on the right-hand side into Equation (I) and solve for

d .

{16 d = - 8 m + 20 m = 4 - 2 d (I) (II)

Substitute

$(I):$ $m = 4 - 2 d$

{16 d = - 8 (4 - 2 d) + 20 m = 4 - 2 d

Distr

$(I):$ Distribute $- 8$

{16 d = - 32 + 16 d + 20 m = 4 - 2 d

AddTerms

$(I):$ Add terms

{16 d = 16 d - 12 m = 4 - 2 d

SubEqn

$(I):$ $LHS - 16 d = RHS - 16 d$

{0 \neq = - 12 m = 4 - 2 d

After substitution and simplification, Equation (I) is a false statement. This means that the system of equations has no solution.

b To graph the system of equations on a coordinate plane, start by rewriting both equations so that they are in slope-intercept form. Consider

m

as the

y -

variable and

d

as the

x -

variable.

{16 d = - 8 m + 20 m = 4 - 2 d (I) (II)

▼

(I), (II):

Write in slope-intercept form

CommutativePropAdd

$(II):$ Commutative Property of Addition

{16 d = - 8 m + 20 m = - 2 d + 4

RearrangeEqn

$(I):$ Rearrange equation

{- 8 m + 20 = 16 d m = - 2 d + 4

SubEqn

$(I):$ $LHS - 20 = RHS - 20$

{- 8 m = 16 d - 20 m = - 2 d + 4

DivEqn

$(I):$ $LHS / (- 8) = RHS / (- 8)$

{m = - 2 d + 2.5 m = - 2 d + 4

The equations have the same slope,

- 2,

but they have different

y -

intercepts. Use the values of the slopes and

y -

intercepts to graph both equations.

The lines are parallel. Since they do not intersect, there is no solution to the system of equations. Vincenzo's team was getting closer and closer to the dark hole when, suddenly, he woke up. Wow, what a cool dream he had tonight!

Closure

Finding the Number of Books and Movies

Vincenzo spends a lot of his free time reading books and watching movies about space travel, distant galaxies, and rocket science.

He counted that he has watched or read

27

things related to space in total, and that the number of movies is

9

more than the number of books. What are the numbers of movies and books about space that Vincenzo has watched or read?

Hint

Write two equations that describe the total number of movies and books about space that Vincenzo has watched or read. Then solve the system of equations by using the Substitution Method.

Solution

Let

b

be the number of books and

m

be the number of movies about space that Vincenzo has read or watched. It is given that the total number of books and movies is

27 .

This means that the first equation can be written by adding

b

and

m

and setting the sum equal to

27 .

b + m = 27

The number of movies is

9

more than the number of books. Using this information, a second linear equation can be written.

m = b + 9

These two equations form a system of equations.

{b + m = 27 m = b + 9

Since

m

is already isolated in Equation (II), solve the system by using the Substitution Method.

{b + m = 27 m = b + 9 (I) (II)

Substitute

$(I):$ $m = b + 9$

{b + b + 9 = 27 m = b + 9

AddTerms

$(I):$ Add terms

{2 b + 9 = 27 m = b + 9

SubEqn

$(I):$ $LHS - 9 = RHS - 9$

{2 b = 18 m = b + 9

DivEqn

$(I):$ $LHS / 2 = RHS / 2$

{b = 9 m = b + 9

Substitute

$(II):$ $b = 9$

{b = 9 m = 9 + 9

AddTerms

$(II):$ Add terms

{b = 9 m = 18

This means that Vincenzo has read

9

books and watched

18

movies about space.

Solving System of Equations Algebraically

Catch-Up and Review

Books and Movies About Space

Substitution Method of Solving a System of Equations

Substitution Method

First Time Spacewalking

Answer

Hint

Solution

Distances Between Planets

Answer

Hint

Solution

Elimination Method

Pit Stop on Exosia

Answer

Hint

Solution

Discovering a New Galaxy

Answer

Hint

Solution

Checking the Solutions of Systems of Equations

Practice Solving Systems of Linear Equations

Numbers of Solution to Systems of Equations

Stars and Planets in the Stellar Nebula

Answer

Hint

Solution

Exploring a Black Hole

Answer

Hint

Solution

Finding the Number of Books and Movies

Hint

Solution

Solving System of Equations Algebraically

Recommended exercises

	14 Theory slides
	11 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions