Algebra 1 > Notes > Yorkcounty Final > Yc > Solving Systems Of Equations

Solving Systems of Equations Lesson 22

Until this point you have only graphed one line at a time. In this unit you will look at a system of two lines and graph them both simultaneously.

y

x

The “solution to a linear system” is the coordinates of the point where the two lines intersect. You will learn four ways to find the solution over the next few lessons: graph and guess, calculator, elimination method, and substitution method.

Graphing to Solve To solve a linear system by graphing and guessing you must: 1) graph each equation on the same set of axes 2) find the coordinates of the point where the two lines intersect

Solve the system 3x - 2y =4 and x + y = 3 by graphing.

Solution: First we must solve each equation for y=mx+b form so we can graph them.

y

3x - 2y = 4 - 2y = -3x + 4 y = -3x + 4 -2 y = 3/2 x - 2 x+y=3 y = -x + 3

The lines look like they intersect at the coordinates of (2,1). But how can we be sure that this is correct. Could it be something more like (2.3, 1.1)? Let’s use a calculator to confirm.

x

To find the intersection of two lines on a TI-83 Graphing Calculator: 1)

Make sure both equations are solved for y and in y=mx + b form.

2)

Press the “y=” button on the top left screen of your calculator

3)

Type both equations in the calculator with one equation at y1 and the other at y2.

4)

Press “zoom” “6” to graph them on a standard graph

5) menu

Press the “2nd” button and then the “trace” button for the Calc

6)

Choose the option that says intersect

7)

Press “enter” three times to get the coordinates where the lines intersect

Try this procedure with the two linear equations on the previous screen to confirm that the solution is (2,1).

Remember that you can enter equations in the calculator that aren’t solved for y=mx + b by noting the steps you would complete to solve them: Example: How would you enter -3x + 9y = 27? If you were solving for y= you would first +3x on both sides and then divide both sides by 9. So enter in the calculator (27 + 3x)/9 at y1.

Checking the System 3x - 2y =4 and x + y = 3 You found the answer to the system to be (2,1). If you substitute the answer into both equations it should produce true statements. 3x – 2y = 4 Substitute (2,1) x + y = 3 3(2) – 2(1) = 4 (2) + (1) = 3 6–2=4 3=3 4=4 Both equations produce true statements, thus the solution to the system is true.

Remember To enter equations into the calculator you must have the equation solved for y. Example: 6x – 2y = 12 You must solve for y to plug the equation into the calculator. “Get y by itself” When you solve for y you should get y = -3x – 6 ; that equation is ready for the calculator.

Solve by Graphing y = 2x and

x + y = 12

Enter both equations in the calculator and use it to solve the system.

Solve by Graphing y = 2x and

x + y = 12

What you enter into the calculator

y = 2x and

x + y = 12

enter in the calculator: y1 = 2x y2 = (12-x) solution: (4,8)

“The solution to a linear system of equations is the coordinate point (x, y) where the two lines intersect.”

Solving Using Elimination ELIMINATE A VARIABLE: EITHER X OR Y When is it smartest to use the elimination method? When the equations are both in general form: ax + by = c.

Solve the system 2x - y = 1 and x + y = 8 using the elimination method: A) (3,5)

B) (3,6)

C)(3,-1)

D) (5,4)

Solution:

To use elimination start by 1) stacking up the equations with matching variables vertical to each other. Indicate that you are going to add them together. 2x - y = 1 + x

+y = 8

This is why some books call this method Linear Combination: The two lines are combined. What we want to happen is one variable to cancel out--We don’t care which. If you were to add vertically, would one of the variables cancel out and make zero?

If you said yes, then you are right. The -y and the +y make 0. 2x - y = 1 +x

+y = 8

3x + 0 = 9

The equations have been added

We now have an equation with only one variable in it. We will solve to find out what x equals when the two lines intersect. 3x = 9 x = 3 The two lines intersect when the x value is 3, but what about the y-value? ( 3, ?) We can substitute the x value of 3 into either equation to find out what y value works? I will use x + y = 8 because that is an easier one to solve: x+y=8 3+y=8 y=5

So, the two lines intersect at (3,5). Use your TI-83 calculator to check this solution: Enter the Equations The equation

Entered in y= menu

x+y=8

y = (8-x)

2x - y = 1

y = (1 - 2x)/-1

Press zoom 6 for a standard 10x10 grid graph Press 2nd Trace to Calc Press the option that says intersect Press enter three times The solution should read x = 3, y = 5

In both of the previous problems, the variable we eliminated had the same coefficient, but opposite signs so they would cancel out and make zero. What if you get a set of equations which is not this way? For Example:

3x + 2y = 8

and

2x - y = 3

If we stack them up: 3x + 2y = 8 2x -

y = 3

We can see that the x’s nor the y’s cancel out. However we can make them cancel out by using the rules of solving equations. We are allowed to multiply both sides of the equation by any number. We will solve the problem on the next slide.

Solve the system of linear equations: 3x + 2y = 8

and

2x - y = 3

3x + 2y = 8 2x -

y = 3

Notice that the coefficients of y are opposite signs, but don’t have the same coefficient, so they currently add to -1y instead of 0. If we could make the bottom equation have a -2y, they would add up to zero. Let’s multiply the bottom equation by a value of 2 to make that happen. 3x + 2y = 8 2( 2x -

y = 3)

The problem becomes 3x + 2y = 8 4x - 2y = 6

Now we can add them:

7x + 0 = 14 Solve for the remaining variable 7x

= 14

x

=2

Now that we know the x-value of the solution is 2, we need to find the yvalue. Do that by the same method as before. Substitute the x-value of 2 into one of the original two equations. It doesn’t matter which.

x=2

3 x + 2y = 8 3(2) + 2y = 8 6

+ 2y = 8

6 - 6 + 2y = 8 -6 2y = 2 2

2

y=1 Thus the solution is (2, 1).

Solve for y

Solving using Substitution You have seen how to solve a system using the graphing approach and using the Elimination approach. You will now learn how to use the method of substitution to solve a system of equations.

Solve the System 4x + 5y = 11 and y = 3x - 13

Since y = 3x - 13, we can substitute 3x - 13 for y in the other equation: 4x + 5y = 11. This gives us: 4x

+

5y

=

4x

+

5(3x-13) =

11 11

Notice there are parentheses around what has been substituted to help us remember to distribute the 5. 4x

+

15x - 65=

11

19x

-

11

65 =

Now we will finish solving for x since there is only 1 variable: 19x

-

65 + 65=

11 + 65

19 x

76

=

19 x

19 =

4

Now we need a y-coordinate for our solution: (4, ?) We must substitute our x value of 4 in one of the two original equations: 4x + 5y = 11 and y = 3x - 13

I will choose the second equation y = 3x - 13 because it is less work to substitute: y = 3x - 13 y = 3(4) - 13 y = 12 - 13 y = -1

( 4, -1 )

Solve the system x = 2y and 4x + 2y = 15 by using the substitution method.

Solution: Since x = 2y we can substitute 2y for the x in the other equation: 4x

+ 2y = 15

4(2y) + 2y = 15 8y

+ 2y

= 15

10y

= 15

10

10

y

= 1.5

Finish solving for y

Now re-substitute to find the value for x: (?, 1.5)

Although we could substitute our y=1.5 in either equation to solve for x. Clearly one of them is easier to use: x = 2y and 4x + 2y = 15 We will use x = 2y: x = 2y x = 2(1.5) x=3 So, the solution is (3, 1.5). This is where the two lines intersect. Don’t forget you can check it in your calculator: Let’s review how:

Solve the following system: y = 3x - 8

and

y=4-x

Solution: Since they are both already solved for y we can set them equal to each other. 3x - 8 = 4 - x This equation has only one variable and can be solved: 3x + x - 8

=

4-x+x

4x - 8

=

4

4x - 8 + 8

=

4+8

4x

=

12

4x/4

=

12/4

x

=

3

Now we must find the value of y:

( 3, ?)

y = 3x – 8 , x = 3 , y = 3(3) – 8 = 9 – 8 = 1 y = 1 Solution (3,1)

Example 1 The length of a rectangle is twice the width. The perimeter is 30 cm. What is the length of the rectangle.

Practical Problems Sometimes you want to use a system of equations to solve practical problems.

Now we have two equations: L = 2W

and

30 = 2L + 2W

It is not necessary to always have x and y as the variables. In fact, when solving a problem, variables that are connected to the words they represent are much smarter. So, when it is solved our answer immediately has meaning. What method would be best used to solve this system? Elimination or Substitution? If you think substitution, you are right since the first equation is solved for L. We can substitute 2W for L in the second equation: 30 =

2L

+ 2W

30 = 2(2W) + 2W

Substituting 2W for L since L=2W

30 =

This equation has 1 variable, solve it…

4W + 2W

30 = 6W 5 = W

The width of the rectangle is 5

We know the width is 5 cm. Because of the first sentence of the problem we have the equation: L = 2W If we substitute 5 for W we get: L = 2 (5cm) L = 10 cm So, the length of the rectangle is 10 cm. Checking the answer: We were told the perimeter is 30cm. See if the Length and Width we discovered by solving works: 10cm

P = 10 + 10 + 5 + 5 P = 30 cm 5 cm

5 cm

10cm

You Try It The perimeter of a rectangle is 40 ft. The length of the rectangle is 4 ft less than three times the width. Find the length.

Solution:

40 = 2L + 2W L = 3W - 4

Solving via Substitution method 40 = 2L

+

40 = 2(3W-4)

2W + 2W

40 = 6W - 8 + 2W 40 = 6W - 8 40 + 8 = 8W - 8 + 8 48 = 8W 48/8 = 8W / 8 6=W If W = 6ft then L = 3W-4:

L = 3(6) - 4

=

14ft

You Try It In a basketball game, Jill and Frank made 30 baskets. Jill scored 5 times as many baskets as Frank. How many baskets did Jill score?

Jill and Francine made 30 baskets: J = # of baskets made by Jill F = # of baskets made by Frank First equation:

J + F = 30

Jill scored 5 times as many baskets as Frank Jill’s baskets are equal to 5 times Frank's baskets Second equation: J = 5F Solve via Substitution: J + F = 30 5F + F = 30

Substitute 5F for J

6 F = 30 F=5 J = 5F and since F = 5, we can substitute J = 5(5) = 25

Now you Try It: The cost of an adult ticket to a baseball game was $1.75. The cost of a student ticket was $1.25. The number of student tickets sold was twice the number of adult tickets. The total income from the sale of tickets was $850. How many student tickets were sold?

Student tickets sold equals twice the number of adult tickets: s = 2a Ticket sales were $850 when student tickets were $1.25 each and adult tickets were $1.75 each: 850 = 1.25s + 1.75a Solving by using the substitution method 850 = 1.25s + 1.75a Substitute 2a for s since s = 2a. 850 = 1.25(2a) + 1.75a 850 = 2.5a + 1.75a 850 = 4.25a 4.25

4.25

200 = a

So, 200 adult tickets were sold, but the question asks for student tickets…..

Since we have the formula s = 2a where “a” represents the number of adult tickets, we can substitute our value of a=200 to find the value of s. S=2a s = 2(200) s = 400

So, 400 adult tickets were sold.