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Solving Systems of Equations by Substitution

Lesson 6.2 (pages 390-396)
by

Rob Frederick

on 15 November 2011

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Transcript of Solving Systems of Equations by Substitution

LINEAR EQUATIONS
BY
SUBSTITUTION
SOLVING SYSTEMS
OF
x + 2y = 6
x - y = 3
Bell Ringer
1.) A swimming pool charges an annual $75 membership fee, and it costs $1.50 each time a member brings a guest. Which equation shows the yearly cost y in terms of guests g?
A.) y = 75g + 1.5
B.) y = -1.5g + 75
C.) y = 1.5g + 75
D.) y = 1.5g + 75g

2.) Which expression represents the phrase "four times the sum of a number and 2"?
A.) 4n + 2
B.) 4(2n)
C.) 4(n + 2)
D.) 4 n + 2
REMEMBER...
last week we
graphed systems.
Sometimes it's not easy to identify a solution by graphing a system of equations.
TODAY...
we will talk about substitution
as an alternative to graphing.
run
SO...
NOW...
practice substitution.
Solve the system by substitution.
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Write it down
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CHECK...
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ASSIGNMENT
Lesson 6.2
Worksheets
A
B
C
Practice
&
,
,
,
This is due on Friday...
We will work together on SOME
of it tomorrow in class...
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So we need alternatives:
Substitution
Elimination
Solve the system:
{
y = 2x
y = x + 5
y = x + 5
y = 2x
Notice that if we graph this system, the lines will cross and there is a solution, but we need to extend our graph to find it.
x
y
So, let's try solving using substitution.
y = 2x
y = x + 5
Remember first what substitution means...
Substitution
If

Then
x = 4
2x = 2(
4
) = 8
So if a variable equals something (no matter what it is: a number, another equation, another variable), we can "plug it in" or substitute it for that variable.
So what can we
substitute here?
SOLVING BY SUBSTITUTION
y = 2x
y = x + 5
(2x) = x + 5
SOLVING BY SUBSTITUTION
Step 1: Solve for one variable
Luckily these are both solved for y already.
Step 2: Substitute
Now, we should be able to solve for
one variable. In this case, x.
SOLVING BY SUBSTITUTION
y = 2x
y = x + 5
(2x) = x + 5
-x -x
x = 5
Step 3: Solve for the first variable
_
_
Now we can use the first variable to solve for the second. (Use the value of x to solve for y).
SOLVING BY SUBSTITUTION
y = 2x
y = 2(5)
y = 10
y = x + 5
(2x) = x + 5
-x -x
x = 5
Step 4: Substitute the first variable
_
_
SOLVING BY SUBSTITUTION
y = 2x
y = 2(5)
y = 10
y = x + 5
(2x) = x + 5
-x -x
x = 5
Step 5: Write the ordered pair and check.
_
_
(5, 10)
The five steps for solving with substitution.
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1.) Solve for one variable in at least one equation, if necessary.
2.) Substitute the resulting expression into the other equation.
3.) Solve that equation to get the value of the first variable.
4.) Substitute that value into one of the original equations and solve.
5.) Write the values from steps 3 and 4 as an ordered pair, (x, y), and check.
OR...
Five steps: the short version (for your HW)...
1.) Solve for one variable.
2.) Substitute into the other equation.
3.) Solve.
4.) Substitute the first variable.
5.) Write as an ordered pair and check.
{
2x + y = 5
y = x - 4
Step 1:
Step 2:
Step 3:
Step 4:
Step 5:
2x + y = 5
y = x - 4
y = x - 4

2x + (x - 4) = 5
2x + (x - 4) = 5
3x - 4 = 5
+ 4 +4
3x = 9
3 3
x = 3
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_
y = (3) - 4
y = -1
(3, -1)
2x + y = 5 y = x - 4
2(3) + (-1) = 5 (-1) = (3) - 4
6 + -1 = 5 -1 = 3 - 4
5 = 5 -1 = -1
MORE PRACTICE...
Solve by substitution.
{
4y - 5x = 9
x - 4y = 11
Step 1: Solve for one variable.
Which equation/variable would be easier to solve for?
Do you want to deal with fractions?

Do you have to?
What was step 2 again?
Step 2: Substitute
You should have gotten



on step 1.
x = 11 + 4y
Now what do we substitute and where?
Step 3: Solve
Step 2 should have given you
4y - 5(11 + 4y) = 9
So what are we solving for?
And how?!
Now, once we've solved...
what do we do with the solution?
Step 4: Substitute
After step 3 we should have gotten
y = -4
What do we do with this information?
Step 5: Rewrite and Check
After step 5 we now have the solution
(-5, -4)
BUT DON'T FORGET TO CHECK YOUR WORK!
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(
-5
,
-4
)
CHECK...
4
y
- 5
x
= 9
4(
-4
) - 5(
-5
) = 9
-16 - (-25) = 9
-16 + 25 = 9
9 = 9

x
- 4
y
= 11
(
-5
) - 4(
-4
) = 11
-5 - (-16) = 11
-5 + 16 = 11
11 = 11
Checking your work takes time, BUT...
it means never having to get a solution wrong.
APPLICATION
One high speed internet provider (ISP) has a $50 setup fee and costs $30 per month. Another provider has no setup fee and costs $40 per month.

Which plan is better?

After how many months will the cost for both providers be the same?
First,
We need to get an equation for both ISP's.
Option 1:
Option 2:
t = 50 + 30m
t = 40m
Let m be months of service and t be total cost.
Now...
Solve using substitution.
t = 50 + 30m
t = 40m
t = 40(5)
t = 200
40m = 50 + 30m
-30m = - 30m
10m = 50
10 10
m = 5
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(5, 200)
THEN...
We interpret the results
(5, 200)
Remember we let m represent months of service and t represent total cost.
SO...
The plans will cost the same amount

after 5 months...

and at that point $200 will have been spent.
So which plan is better?
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