Transcript Ch 7.2 Linear Systems by Substitution

Algebra 1
Ch 7.2 – Linear Systems by
Substitution
Objective
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Students will solve systems of linear
equations using the substitution
method.
Before we begin…
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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
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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
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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
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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
Equation # 1
+x
y= x+1
Now that I have isolated the value of y, I will use the expression
x + 1 and substitute it into the 2nd equation for y.
Step 2
Equation #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
2x + x + 1 = -2
3x + 1 = -2
-1
3x
3
-1
= -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
x = -1
Revised equation from step 1
y= x+1
Substitute & Solve
y = -1 + 1
Solution
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
2x + y = -2
-(-1) + 0 = 1
2(-1) + 0 = -2
1=1
-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
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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
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On the next couple of slides are some practice
problems…The answers are on the last slide…
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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…
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If you cannot find the error bring your work to
me and I will help…
Your Turn
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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
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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
and
y=x+7
x – 2y = -25
and
3x – y = 0
x–y=0
and
12x – 5y = -21
y = 3x
and
x = 3y
7g + h = -2
and
g – 2h = 9
Your Turn Solutions
Answers can vary
1. Solve for y in the
2nd equation
2. Solve for m in the
2nd equation
3. Solve for d in the
2nd equation
4.
5.
6.
7.
8.
9.
10.
(6,2)
(3,8)
(2,9)
(5,15)
(-3,-3)
(0,0)
(1/3, -4 1/3)
Summary
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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
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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:
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Have your name, date and period as well a lesson number
as a heading.
Do each of the your turn problems showing all work
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Have a 1 paragraph summary of the lesson in your own
words
Please be advised – I will not give any credit for work
submitted:
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Without a complete heading
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Without showing work for the your turn problems
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Without a summary in your own words…
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