Linear Equations Systems of Linear Equations - Introduction of Linear Equations.pdfLinear Equations Systems of Linear Equations - Tables Objectives: • Solving systems of Linear Equations

Linear EquationsSystems of Linear Equations - Introduction

Objectives:• What are Systems of Linear Equations• Use an Example of a system of linear equations

If we have two linear equations, 𝑦 = 𝑥 + 2 and 𝑦 = 3𝑥 − 6, can these two equations equal one another?

𝑦 = 𝑥 + 2

𝑦 = 3𝑥 − 6

x f(x)=x+2 f(x)=3x-6

2 4 0

4 6 6

6 8 12

So what does it mean when two linear equations equal one another?

The equations must intersect.



Why do we care?

PRICE

QUANTITY

Demand

As the demand of your product goes down, the quantity of product in your stores goes up. In order to sell your product, the price must go down.

SupplyAs your product supply goes up, your cost to produce your product goes up.

Imagine you have a company selling your favorite product.

Where these two lines intersect, is the price you should charge to break even.

Can you see the inequality possibilities?



A farmer has 100 animals consisting of cows, pigs, and chickens. The farmer takes these animals to market and sells them for $100. If the farmer sells the cows for $10, the pigs for $3 and sells the chickens 2 for $1, how many of each animal does he sell if there are no partial animals sold?

What do we know?

Let C = # of cows, let P = # of pigs, and S = # of chickens.

Total number of animals sold: C + P + S = 100

Total cost of the animals sold: 10C + 3P + 0.5S = 100

We have two equations, but three variables.

The number of each animal sold must be greater than one and have no fractions.



Many times we can solve for one variable and then substitute that expression into a second equation. There cannot be many cows, so lets solve an equation in terms of S.

C + P + S = 100; then S = 100 – C – P.

Substituting: 10C + 3P + 0.5(100 – C – P) = 100

10C + 3P + 50 – 0.5C – 0.5P = 100

9.5C + 2.5P = 50

19C + 5P = 100

P = -19

5C +20; this y = mx + b

If we think about this, the number of cows can not be zero, why?

The number of cows must be a multiple of 5, why? Can not be greater than 10, why?

Therefore, the number of cows must be? 5



Knowing one variable in our three variable system of linear equations means we now have two equations and two variables. We can now solve the other two variables.

C + P + S = 100 or 5 + P + S = 100 or P + S = 95

50 + 3P + 0.5S = 100 or 3P + 0.5S = 50

1) P + S = 952) 3P + 0.5S = 50 If we multiply the first equation by 3;

1) 3P + 3S = 2852) 3P + 0.5S = 50

Remember, these equations must be equal; therefore, subtracting equation 2 from 1 …

2.5S = 235 so S = 94

This means, the farmer sold 5 cows, 94 chickens, and 1 pig.

Linear EquationsSystems of Linear Equations - Tables

Objectives:• Solving systems of Linear Equations using tables.

A set of linear equations that has more than one variable is called a system of linear equations. The single pair of variables that satisfies both equations is their unique solution.

One method to solve a system of linear equations is to make a table of values for each equation and compare each table for the common solution of both equations.

Lionel is x years old and his sister is y years old. The difference in their ages is 1 year. The sum of 4 times Lionel's age and his sister’s age is 14 years. Find Lionel’s and his sister’s ages.

We have two equations: x – y = 1 and 4x + y = 14.

x 2 3 4

y 1 2 3

x – y = 1 4x + y = 14

x 1 2 3

y 10 6 2

In the second table, the first two pairs does not make sense, because Lionel is older than his sister. Lionel’s age must be 3 years and his sister’s age must be 2 years.

Linear EquationsSystems of Linear Equations - Tables

How can we use a calculator to solve systems of Linear Equations?

Given the two equations : 8x + y = 38 and x – 4y = 13; find the solution of this system.

TI-84

2nd TBLSET / Tblstart = 0 / Tbl =1 / Ind:auto Dep:auto

Solve each equation for y; input into Y2 and Y3

2nd TABLE / arrow down until Y values are equal

TI-73

2nd TBLSET / Tblstart = 0 / Tbl =1 / Ind:auto Dep:auto

Solve each equation for y; input into Y2 and Y3

2nd TABLE / arrow down until Y values are equal

The unique solution for this system of linear equations is x = 5; y = -2.

Bookwork: Math in Focus – Practice 5.1; page 196, problems 1-12.

Objectives:• Solving systems of Linear Equations using tables.

Linear EquationsSystems of Linear Equations - Elimination

Objectives:• Solving systems of Linear Equations using Elimination.

When equations become more complicated, the use of tables may not be the best method to solve the system.

Lets look at two equations: 𝑥 + 𝑦 = 8

𝑥 + 2𝑦 = 10

Using Bar Models:

x y = 8

x yy = 10

The difference in length of the two bars is 2 units; therefore, y = 2 because the one y-bar must equal the difference of 2 units.

If; x 2 = 8

Then x must equal 6 units.



What did we do to solve the previous system of linear equations?

We eliminated one variable to solve for the other variable, then substituted the known variable into one of the equations to solve for the eliminated variable.

𝑥 + 𝑦 = 8 𝑥 + 2𝑦 = 10

If both equations have a variable with the same coefficient, elimination of that variable by subtraction will yield one equation with only one variable.

𝑥 + 2𝑦 − 𝑥 + 𝑦 = 10 − 8

𝑥 + 2𝑦 − 𝑥 − 𝑦 = 2

𝑥 − 𝑥 + 2𝑦 − 𝑦 = 2

𝑦 = 2

Substituting y = 2 into the first equation yields x = 6.

The solution is then given by x = 6, y = 2



We know that addition is the opposite of subtraction. Given the following equations:

4𝑥 + 𝑦 = 9 3𝑥 − 𝑦 = 5

We can eliminate y by adding the two equations:

(4𝑥 + 𝑦) + (3𝑥 − 𝑦) = 9 + 5

4𝑥 + 3𝑥 + 𝑦 − 𝑦 = 14

7𝑥 = 14

𝑥 = 2, 𝑦 = 1

We were able to eliminate a variable by using addition.

This method works well when a variable has the same coefficient in both equations.



What can we do when there is no common coefficient?

2𝑥 + 3𝑦 = 7 𝑥 + 6𝑦 = 8

Using Bar Models we see:

x x y y y = 7

x y y y y y y = 8

By using a copy of the second bar model we can redraw the equations:

x x y y y = 7

x x y y y y y y yyy y yy = 16

By using elimination, y = 1. Our solution becomes x = 2, y = 1

We solved this system by multiplying the second equation by 2, giving a common coefficient for x, then using the elimination method.

Bookwork: Math in Focus – Practice 5.2; page 209, problems 1- 9.

Linear EquationsSystems of Linear Equations - Substitution

Objectives:• Solving systems of Linear Equations using Substitution.

You have seen how elimination can simplify two equations with two variables into one equation with one variable. Lets look at our first example again:

𝑥 + 𝑦 = 8 𝑥 + 2𝑦 = 10

x y = 8

x yy = 10

You can redraw the first bar graph to look like:

x = 8 − y

Then redraw the second bar graph to look like:

yy = 108 − y

Solving for y; y = 2, then substituting into the first equation yields x = 6.

Linear EquationsSystems of Linear Equations - Substitution


By solving for one variable in one equation, you can substitute that value into the second equation and solve for the one variable. Lets look at a second example algebraically.

3𝑥 − 𝑦 = 18 𝑦 = 𝑥 − 4

Here, the second equation is already solved for y. If we substitute this value of y into the first equation…

3𝑥 − 𝑥 − 4 = 18

3𝑥 − 𝑥 + 4 = 18

2𝑥 + 4 = 18

2𝑥 = 14

𝑥 = 7 ; 𝑦 = 3

Question: could you solve for x in the second equation and substitute for y in the first equation?

Bookwork: Math in Focus – Practice 5.2; page 209, problems 10 - 18.

Linear EquationsSystems of Linear Equations - Choose

Objectives:• Solving systems of Linear Equations using - Choose the method.

You have seen that both methods, elimination and substitution, can be used to solve a system of linear equations.

Both methods find the same solution set, don’t they? This is because two lines can only intersect at one point; therefore, there is only one solution.

In Algebra II you will see multiple solutions because Algebra II has functions with exponents which changes the line into a curve.

Which method is the best method?

That depends on which method you like to use and what the equations look like.

Would you use substitution for the system; 3𝑝 + 2𝑞 = 1 𝑎𝑛𝑑 2𝑝 − 5𝑞 = −12 ?

No, because whatever variable you solved for, there would be fractions.

Bookwork: Math in Focus – Practice 5.2; page 209, problems 19 – 25.

Linear EquationsInconsistent and Dependent Systems of

Linear EquationsObjectives:• Understand and identify inconsistent systems of linear equations• Understand and identify dependent systems of linear equations

You have seen that given two linear equations, the unique solution is one point.

What happens if a system does not have a unique solution?

Given the following system of linear equations:

2𝑥 + 𝑦 = 14𝑥 + 2𝑦 = 4

If we multiply the first equation by 2 and solve by elimination…

4𝑥 + 2𝑦 = 24𝑥 + 2𝑦 = 4

We get 0 = 2; this is a false statement; meaning this system does not have a solution.

Lets look at the coefficients of these equations; 2, 1, and 1 and 4, 2, and 4.

Notice the ratios of these coefficients. A and B are the same ratio; however, C is a different ratio.


Linear EquationsObjectives:• Understand and identify inconsistent systems of linear equations• Understand and identify dependent systems of linear equations

Lets look at the slopes of these equations: y = -2x + 1 and y = -2x + 2.

The slopes are the same; meaning the lines are parallel.

Parallel lines do not intersect; therefore, there is no unique solution.

A linear system that does not have a unique solution is said to be inconsistent or has no solution.


Linear EquationsLets consider the following system of linear equations:

𝑥 + 2𝑦 = 22𝑥 + 4𝑦 = 4

If we use substitution to solve this system…

𝑥 = 2 − 2𝑦 from equation 1.

Substituting this into equation 2 yields, 2 2 − 2𝑦 + 4𝑦 = 4

Simplifying yields 4 = 4. This is always true, indicating that all values of x and y are solutions.

What happens when you graph the two equations? Solving for y in both equations

yields 𝑦 = −1

2𝑥 + 1. These are the same equation resulting in the same line;

therefore all points are a solution.

Looking at the coefficients in the two equations; 1, 2, and 2 and 2, 4, and 4, the ratios of A, B, and C are the same. This means the equations are the same, doesn’t it?

A system of equations that have an infinite number of solutions, or has the same slope and same y-intercepts, or if one equation is a multiple of another, the system is a dependent system or has infinite solutions.

Bookwork: Math in Focus; page 234, problems 1 – 14.

Linear Equations Systems of Linear Equations - Introduction of Linear Equations.pdfLinear Equations Systems of Linear Equations - Tables Objectives: • Solving systems of Linear Equations

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