Download Solving Systems by Graphing

Solving Systems by
Substitution
Unit IX, Lesson 2
Online Algebra 1
VHS@pwcs
System Review
 Remember that we can solve a system many
ways, but in Algebra 1, we are going to solve
them the following ways.



Graphing
Substitution
Elimination.
 Also recall that there are three types of
solutions for systems.



One solution: an ordered pair
No solution
Infinite solutions.
Solving Systems by
Substitution.
 Recall that if we have the following
expression, 3x – 7 and we know that x =
5, we can substitute 5 in for x.
 So the value of that expression is:



3(5) – 7
15 – 7
8
 We can use substitution to solve
systems.
Solving Systems by
Substitution.
y = 2x
6x – y = 8
6x – 2x = 8
4x = 8
x=2
y = 2(2)
y=4
1. Since the first equation tells us that y
= 2x, we can substitute 2x in for y in
the second equation.
2. Solve the second equation for x.
 Combine the like terms
 Divide both sides by 4
3. Now sub 2 in for x in one of the
equations to solve for y.
 Using the first would be the easier
since it is already solved for y
4. Our solution is (2, 4)
Checking our solution.
y = 2x
6x – y = 8
We can check by substituting our ordered pair into
BOTH equations and simplifying.
The ordered pair (2, 4) must work for both
equations!
Y = 2x
6x – y = 8
4 = 2(2)
6(2) – 4 = 8
4=4
12 – 4 = 8
8=8
The ordered pair (2, 4)
works for both
equations so we know
that it is the solution to
the linear system.
Let’s try another!
3x – 6y = 30
1.
y = -6x + 34
3x – 6(-6x + 34) = 30
2.

3x + 36x – 204 = 30

39x – 204 = 30

39x = 234
x=6

3.
Y = -6(6) + 34
4.
Get rid of the parentheses using the
distributive property
Combine like terms
Add 204 to both sides
Divide by 39
Substitute 6 in for x in the second equation
and solve for y.

Y = -36 + 34
Y = -2
Since the second equation is solved for y,
we can substitute -6x + 24 in for y in the first
equation.
Solve for x.
Again this is easier since it is already
solved for y.
Our solution is (6, -2)
3x + 5y = 2
Another!
 This time neither equation is solved for a
variable, so we solve one of the equations
for a variable.
 Since the x in the second equation has no
coefficient it is easier to solve that for x.

Subtract 4y from both sides.
 Now that the second is solved for x, we can
sub -4 – 4y in for x in the first equation.

It must be the first since it is the second
that we solved.
 Now follow the steps in the last two slides to
solve!
 Our solution is (4, -2)
x + 4y = -4
x + 4y – 4y = -4 – 4y
x = -4 – 4y
3(-4 – 4y) + 5y = 2
-12 - 12y + 5y = 2
-12 + -7y = 2
-7y = 14
y = -2
x = -4 – 4(-2)
X = -4 + 8
X=4
Try this: Click for the steps!
2x + y = 5
2y = 10 – 4x
2x – 2x + y = 5 – 2x
y = 5 – 2x
2(5 – 2x) = 10 – 4x
10 – 4x = 10 – 4x
10 + 4x – 4x = 10 – 4x + 4x
10 = 10
So there are an infinite
number of solutions.
 Let’s solve the first for y; it’s
easier!
 We need to sub 5 – 2 in for y
into the second equation and
solve for x.
 Notice that when we try to get
the x’s all on one side they
cancel out and we get 10 = 10.


Remember, from Unit 4 that
this means that our solution is
all real numbers.
If we had gotten a false
statement there would be no
solution.
On your own!
Homework
page 278:
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