Algebra 1

Ch 7.2 – Linear Systems by Substitution

Objective

 Students will solve systems of linear equations using the substitution method.

Before we begin…

    In the last lesson we solved systems of equations by graphing… In this lesson we will solve equations algebraically by substitution method… Recall that you cannot solve an equation with two variables….

In the substitution method what you will do is solve one of the equations for one of it variables and then use that phrase to substitute into the other equation, thereby creating an equation with one variable that you can solve…

Process

 1.

2.

3.

4.

The process to solve a system of equations is as follows: Solve one of the equations for one of its variables.

Substitute the expression from step 1 into the other equation and solve for the other variable.

Substitute the value from step 2 into the equation from step 1 and solve Check the solution in each of the original equations

Note: This is a multi step process…to be successful you must be organized!

Comments

    Before we look at an example, keep in mind… You will have to make decisions here… That means you will have to analyze each equation first… Then choose an equation that you can easily and quickly solve for one of its variables, then follow the process…

Example

 Solve the linear system -x + y = 1 2x + y = -2 Equation #1 Equation #2 In this system of equations it will be easy to solve either equation for y My decision will be to work with equation # 1 My reason for making that decision is that x has no coefficient and I only have to add x to both sides and I have solved the equation for y Let’s see what that looks like….

Step 1

Solve 1 equation for one of its variables -x + y = 1 +x +x y = x + 1 Equation # 1 Now that I have isolated the value of y, I will use the expression x + 1 and substitute it into the 2 nd equation for y.

Step 2

Substitute the expression from step 1 into the other equation and solve for the other variable In this example I will substitute x + 1 from equation #1 for y Then solve the equation for x 2x + y = -2 Equation #2 2x + x + 1 = -2 3x + 1 = -2 - 1 -1 3x = -3 3 3 x = -1

Step 3

Substitute the value from step 2 into the revised equation from step 1 and solve Value from step 2 Revised equation from step 1 Substitute & Solve Solution x = -1 y = x + 1 y = -1 + 1 y = 0 Algebraically, the solution to the system of linear equations is the ordered pair (-1, 0)

Step 4

Check the solution in each of the original equations

Solution, x = -1, y = 0 Ordered pair (-1, 0)

-x + y = 1 -(-1) + 0 = 1 1 = 1 2x + y = -2 2(-1) + 0 = -2 -2 = -2 True True

Since the ordered pair (-1, 0) make both of the statements true, the solution to the system of equations is (-1, 0)

Comments

   Again…these are multi-step processes. To be successful you must be organized and lay out your work step by step.

Additionally, at this stage you should have mastered rules for working with integers (positive & negative numbers) and fractions… If you have not mastered those basic concepts yet…you need to do something about it!

Comments

 On the next couple of slides are some practice problems…The answers are on the last slide…  Do the practice and then check your answers…If you do not get the same answer you must question what you did…go back and problem solve to find the error…  If you cannot find the error bring your work to me and I will help…

Your Turn

 1.

2.

3.

Tell which equation you would use to isolate the variable. Explain your reasoning.

2x + y = -10 3x – y = 0 m + 4n = 30 m – 2n = 0 5c + 3d = 11 5c – d = 5

Your Turn

 4.

5.

6.

7.

8.

9.

10.

Use the substitution method to solve the systems of linear equations y = x – 4 and 4x + y = 26 2c – d = -2 and 4c + d = 20 2x + 3y = 31 x – 2y = -25 x – y = 0 and and and y = x + 7 3x – y = 0 12x – 5y = -21 y = 3x 7g + h = -2 and and x = 3y g – 2h = 9

Your Turn Solutions

Answers can vary 1.

Solve for y in the 2 nd equation 2.

3.

Solve for m in the 2 nd equation Solve for d in the 2 nd equation 4.

5.

6.

(6,2) (3,8) (2,9) 7.

8.

9.

10.

(5,15) (-3,-3) (0,0) (1/3, -4 1/3)

Summary

   A key tool in making learning effective is being able to summarize what you learned in a lesson in your own words… In this lesson we talked about

solving linear systems by substitution.

Therefore, in your own words summarize this lesson…be sure to include key concepts that the lesson covered as well as any points that are still not clear to you… I will give you credit for doing this lesson…please see the next slide…

Credit

  

I will add 25 points as an assignment grade for you working on this lesson… To receive the full 25 points you must do the following:

  Have your name, date and period as well a lesson number as a heading.

Do each of the your turn problems showing

all

work  Have a 1 paragraph summary of the lesson in your own words

Please be advised – I will not submitted: give any credit for work

 Without a complete heading   Without showing work for the your turn problems Without a summary in your own words…