Systems of Linear Equations A Tutorial Designed for Mr. Wilson’s Algebra Classes. Click on my alma mater whenever it appears to continue!

Systems of Systems of Linear EquationsLinear EquationsA Tutorial Designed for Mr. A Tutorial Designed for Mr. Wilson’s Algebra Classes.Wilson’s Algebra Classes.

Click on my alma mater whenever Click on my alma mater whenever it appears to continue!it appears to continue!

ObjectiveObjective

This This interactiveinteractive presentation is used as a presentation is used as a supplement to our unit on solving supplement to our unit on solving systems of linear equations in two systems of linear equations in two variables. After completing this unit, the variables. After completing this unit, the student will be able to employ several student will be able to employ several methods to solve a system of linear methods to solve a system of linear equations, and apply their use in real-equations, and apply their use in real-world examples.world examples.

Michigan Michigan StandardsStandardsOur unit on linear systems addresses the Our unit on linear systems addresses the

state standards of:state standards of:

Mathematics, Strand V:Mathematics, Strand V:

Standard 1, Benchmarks 1,4Standard 1, Benchmarks 1,4

Standard 2, Benchmarks 3,4Standard 2, Benchmarks 3,4

ReviewReview

Before we begin, let’s Before we begin, let’s review some things about review some things about linearity and linear linearity and linear equations…equations…

LinearityLinearity

A A Linear EquationLinear Equation in in 2 variables is of the 2 variables is of the form:form:

ax+by+c=0ax+by+c=0

Notice that this can Notice that this can be arranged into be arranged into normal normal slope-slope-interceptintercept form: form:

y=mx+by=mx+b-8-6-4-202468

101214

y=6x

y=2x+4

What is a System?What is a System?

(Click on your answer)(Click on your answer)

A system of linear equations is:A system of linear equations is:

a. A set of parabolasb. A set of two or more linesc. A stereo component

What are What are Solutions?Solutions?

A solution to a A solution to a system of system of equations is any equations is any point where the point where the graphs of the graphs of the lines intersect:lines intersect:

True False-8-6-4-202468

101214

y=6x

y=2x+4

YES!A system of linear equations is a set of two or more lines, grouped by the word “and”, such as y=x+1 and y=x-1.

Try Again

TRUE!TRUE!A solution to a system of A solution to a system of equations is any point that equations is any point that satisfies all of the equations. satisfies all of the equations. Graphically, this is any point Graphically, this is any point of intersection of the system. of intersection of the system. Since we are dealing with Since we are dealing with systems of two linear systems of two linear equations, there can be equations, there can be at at mostmost one solution. Can you one solution. Can you think of why this is?think of why this is?

Try Again

Sample SystemSample SystemExamine our system Examine our system at the right. There at the right. There is one solution to is one solution to our system at our system at (1,6)(1,6), , because this is a because this is a point of intersection. point of intersection.

-1, -6

0, 0

1, 6

2, 12

-1, 2

0, 4

1, 6

2, 8

-8

-6

-4

-2

0

2

4

6

8

10

12

14

y=6x

y=2x+4

Solution to SampleSolution to Sample

We must always verify a proposed We must always verify a proposed solution algebraically. We propose (solution algebraically. We propose (11,,66) ) as a solution, so now we plug it in to both as a solution, so now we plug it in to both equations to see if it works:equations to see if it works:

y = 6xy = 6x andand y = 2x + 4y = 2x + 4,,-6x +y = 0-6x +y = 0 andand -2x + y - 4 = 0-2x + y - 4 = 0,,-6(-6(11)+()+(66)= 0)= 0 andand -2(-2(11) +() +(66) - 4 = ) - 4 =

00..Yes, (1,6) Satisfies both equations!Yes, (1,6) Satisfies both equations!

Other MethodsOther Methods

There are several other methods of solving There are several other methods of solving systems of linear equations. Each is best systems of linear equations. Each is best used in different situations. Choose from the used in different situations. Choose from the list:list:

Substitution Method Elimination Method Matrix Algebra

Substitution Substitution MethodMethod

Substitution method is used when it Substitution method is used when it appears easy to solve for one variable appears easy to solve for one variable in terms of the other. The goal is to in terms of the other. The goal is to reduce the system to two equations of reduce the system to two equations of one unknown each. Let’s examine our one unknown each. Let’s examine our example from earlier:example from earlier:

-2x + y = 4-2x + y = 4

-6x + y = 0-6x + y = 0


Notice that we can arrange the second Notice that we can arrange the second equation to slope-intercept form:equation to slope-intercept form:

y = y = 6x6x

Now we have Now we have yy solved in terms of solved in terms of xx. . The second step is to plug our value The second step is to plug our value for y back into the first equation:for y back into the first equation:

-2x + -2x + 6x6x = 4 = 4


Now we solve the equation for its only Now we solve the equation for its only unknown, x:unknown, x:

-2x + 6x = 4-2x + 6x = 4

4x = 44x = 4

x = 1x = 1

Notice that our x value matches our Notice that our x value matches our earlier value (1, 6).earlier value (1, 6).


Now we plug our Now we plug our xx value back into the first value back into the first equation: equation:

y = 6xy = 6x

y = 6(y = 6(11))

y = 6y = 6

Again, notice that our y value is the same as Again, notice that our y value is the same as what we obtained earlier. Thus, we have what we obtained earlier. Thus, we have found the same point, found the same point, (1,6)(1,6), as a solution to , as a solution to our system.our system.


Now you try. Use the substitution Now you try. Use the substitution method to solve this system:method to solve this system:

5x + 6y = 105x + 6y = 102y = 32y = 3

Your first step is to:Your first step is to:a. Solve the first equation for xb. Solve the second equation for y c. I forgot, I need to review the substitu

tion method


You cannot solve the first You cannot solve the first equation for equation for xx, because you have , because you have no value for the no value for the yy variable. Take variable. Take another look at the problem, another look at the problem, you’ll notice that it is easier to you’ll notice that it is easier to solve the second equation for solve the second equation for yy in in terms of terms of xx..

Substitution Substitution MethodMethodYes, solving the second equation for Yes, solving the second equation for yy

yields:yields:

y = 3/2y = 3/2

The next step is:The next step is:a. Plug y = 3/2 into 5x + 6y = 10b. Plug y = 3/2 into 2y = 3c. I’m not sure, I need to review


We already solved We already solved 2y = 32y = 3 to obtain to obtain

y = 3/2y = 3/2, so we cannot plug this value, so we cannot plug this value

back into the same equation. Go backback into the same equation. Go back

and examine your choices again.and examine your choices again.


Yes, we now plug our Yes, we now plug our yy value value into the first equation:into the first equation:

5x + 6(5x + 6(3/23/2) = 10) = 10

Simplifying, we get:Simplifying, we get:

5x + 9 = 105x + 9 = 10

5x = 15x = 1

x = 1/5x = 1/5

Substitution Substitution MethodMethodWe now have values for each We now have values for each

variable:variable:

x = 1/5x = 1/5 and and y = 3/2y = 3/2So the solution to our system is the So the solution to our system is the

point:point: (1/5, 3/2)(1/5, 3/2)

We know that this can be the only We know that this can be the only solution because two lines can solution because two lines can intersect in at most one point. intersect in at most one point.

TestTest

This review test will help you study This review test will help you study for our unit test on systems of linear for our unit test on systems of linear equations. Once you begin this equations. Once you begin this practice test, you will not be allowed practice test, you will not be allowed to go back and review.to go back and review.

Review Again Take Test

Question 1Question 1

-2

-1

0

1

2

3

4

5

6

-1 0 1 2

How many How many solutions exist for solutions exist for the system at the the system at the right?right?

a. 0b. 1c. 2d. Infinite

YES!YES!The lines are The lines are parallel, so they parallel, so they never intersect. A never intersect. A solution to a solution to a system of system of equations is a point equations is a point where the lines where the lines intersect, thus we intersect, thus we have no solution to have no solution to our system.our system.

-2

-1

0

1

2

3

4

5

6

-1 0 1 2



Solve the system by substitution Solve the system by substitution method:method:

5x – y = -135x – y = -13

2x + 3y = -122x + 3y = -12The solution is:The solution is:a. (-3, -2)b. (-2, 3)c. (13, 4)d. No Solution

YES!YES!The solution is The solution is (-3, -2)(-3, -2). You can . You can verify this by plugging it into the verify this by plugging it into the system:system:

5(5(-3-3) – () – (-2-2) = -13) = -13

2(2(-3-3) + 3() + 3(-2-2) = -12) = -12



Solve the system using elimination method:Solve the system using elimination method:

2x + 5y = 72x + 5y = 7

3x + y = -93x + y = -9

The solution is:The solution is:

a.a. (12, -4)(12, -4)

b.b. (-4, 12)(-4, 12)

c.c. (4, -21)(4, -21)

d.d. No SolutionNo Solution


YES!YES!The solution is The solution is (4, -21)(4, -21). You can verify . You can verify this by plugging it into the system:this by plugging it into the system:

2(2(44) + 5() + 5(-21-21) = 7) = 7

3(3(44) + () + (-21-21) = -9) = -9


Use any method to solve the system:Use any method to solve the system:

-2x + 3y = 10-2x + 3y = 10-2x + 3y = -10-2x + 3y = -10

The solution is:The solution is:a.a. (2, 3)(2, 3)b.b. (2, -3)(2, -3)c.c. (3, -2)(3, -2)d.d. No SolutionNo Solution


YES!YES!Notice from our Notice from our system that both system that both equations have equations have the same slope, the same slope, but different y-but different y-intercepts. Thus, intercepts. Thus, they can never they can never intersect. intersect. Can Can you think of how you think of how they could they could instead have instead have infinite solutions?infinite solutions?

-5.00

-4.00

-3.00

-2.00

-1.00

0.00

1.00

2.00

3.00

4.00

5.00

6.00

-1.00 0.00 1.00 2.00

-2x+3y=10

-2x+3y=-10

Question 5 Question 5 On Your OwnOn Your Own

Question 5 is a story problem that Question 5 is a story problem that demonstrates a real-world use of demonstrates a real-world use of linear systems. Study it carefully, linear systems. Study it carefully, solve the question on your own, solve the question on your own, and think about how powerful and think about how powerful linear systems can be. linear systems can be.


Your company currently uses widgets and Your company currently uses widgets and gadgets to produce your best selling product, gadgets to produce your best selling product, the Ultimate. Looking over your books you see the Ultimate. Looking over your books you see that in May you bought 200 widgets and 400 that in May you bought 200 widgets and 400 gadgets for $500, and in June you bought 250 gadgets for $500, and in June you bought 250 widgets and 250 gadgets for the same cost, widgets and 250 gadgets for the same cost, $500. How much does one widget cost? One $500. How much does one widget cost? One gadget? If a new supplier offered to sell you gadget? If a new supplier offered to sell you widgets for 75% cost of what you currently widgets for 75% cost of what you currently pay, but gadgets would cost 10% more than pay, but gadgets would cost 10% more than what you currently pay, should you switch to what you currently pay, should you switch to this new supplier or stay with your current this new supplier or stay with your current supplier?supplier?


Solutions to question 5 will be Solutions to question 5 will be discussed in class. discussed in class.

Now that you have completed this Now that you have completed this presentation, you have at least presentation, you have at least one use for linear systems one use for linear systems theory.theory.

Can you think of at least 3 other Can you think of at least 3 other examples of systems of linear examples of systems of linear equations?equations?

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Systems of Linear Equations A Tutorial Designed for Mr. Wilson’s Algebra Classes. Click on my alma mater whenever it appears to continue!