Solving Systems by Graphing and Substitution

Solving Systems of Linear Equations by Graphing
y  x 
2 4
1
y   x

3
3






Solution: 3,2
2  234
2  2
 
1
2   3 
3
3
2  2

Graphing to Solve a Linear System
Let's summarize! There are 4 steps to solving a linear system using a
graph.
Step 1: Put both equations in
slope - intercept form.
Step 2: Graph both equations on
the same coordinate plane.
Step 3: Estimate where the
graphs intersect.
Step 4: Check to make sure your
solution makes both equations
true.
Solve both equations for y, so that
each equation looks like
y = mx + b.
Use the slope and y - intercept for
each equation in step 1. Be sure
to use a ruler and graph paper!
This is the solution! LABEL the
solution!
Substitute the x and y values into
both equations to verify the point
is a solution to both equations.

3
y


y x
 4 
1




Solution: 1,3
3 3 3  411
3 3

x y 
3
x  y

 
3 y x   
2 0


Solution: 1,2 
3 3
212  0
0  0


12  3
2yx

CLASSIFICATION OF LINEAR SYSTEMS (p.278)
Classification
Consistent and
Independent
Consistent and
Dependent
Inconsistent
Number of
Solutions
Exactly One Infinitely Many None
Description Different Slopes
Same Slope,
Same y-intercept
Same Slope,
Different
y-intercept
Graph

Y=3x+4
Y=-3x+2
Y=1/2 x + 10
Y = 1/3 x + 10
Y = 4x + 5
Y = 4x + 5
Y = -3x + 1
Y = -3x – 1
2y = 10x + 14
y = 5x +7
Different slopes CONSISTENT and INDEPENDENT
so there is 1 solution to the system
Different slopes
Same slope,
Same y-intercept
Same slope,
Different y-intercepts
CONSISTENT and INDEPENDENT
so there is 1 solution to the system
CONSISTENT and DEPENDENT
So there are infinite solutions
Not slope-intercept form
Change the 1st equation to
Y=5x+7, then Same Slope,
Same y-intercept
INCONSISTENT
So there is no solution
CONSISTENT and DEPENDENT
So there are infinite solutions

Example 1
x + 5y = 9
3x – 2y = 12
9
x
5
y
 

(1)
(2)
To solve, rewrite
each equation in the
form y = mx +b
Isolating y in line (1) Isolating y in line (2)
x + 5y = 9
5y = -x + 9
1 9
5 5
y   x 
3x – 2y = 12
-2y = -3x + 12
3 12
2
x
y
 


3
y  x 
6
2

What type of system is it?
1
5
9
5
m
b
 

1 9
5 5
y   x 
3
y  x 
6
2
What is the slope and
y-intercept for line (1)?
What is the slope and
y-intercept for line (2)?
3
2
6
m
b

 
Since the lines have different slopes they will intersect.
The system will have one solution and is classified as
being consistent-independent.

Objective
The student will be able to:
solve systems of equations using
substitution.
A-REI.3.6

Solving Systems of Equations
 You can solve a system of equations
using different methods. The idea is to
determine which method is easiest for
that particular problem.
 These notes show how to solve the
system algebraically using
SUBSTITUTION.

Solving a system of equations by substitution
Step 1: Solve an equation
for one variable.
Step 2: Substitute
Step 3: Solve the equation.
Step 4: Plug back in to find
the other variable.
Step 5: Check your
solution.
Pick the easier equation. The goal
is to get y= ; x= ; a= ; etc.
Put the equation solved in Step 1
into the other equation.
Get the variable by itself.
Substitute the value of the variable
into the equation.
Substitute your ordered pair into
BOTH equations.

1) Solve the system using substitution
x + y = 5
y = 3 + x
for one variable.
Step 2: Substitute
The second equation is
already solved for y!
x + y = 5
x + (3 + x) = 5
2x + 3 = 5
2x = 2
x = 1

x + y = 5
y = 3 + x
the other variable.
x + y = 5
(1) + y = 5
y = 4
Step 5: Check your
solution.
(1, 4)
(1) + (4) = 5
(4) = 3 + (1)
The solution is (1, 4). What do you think the answer
would be if you graphed the two equations?

Which answer checks correctly?
3x – y = 4
x = 4y - 17
1. (2, 2)
2. (5, 3)
3. (3, 5)
4. (3, -5)

3y + x = 7
4x – 2y = 0
for one variable.
Step 2: Substitute
It is easiest to solve the
first equation for x.
3y + x = 7
-3y -3y
x = -3y + 7
4x – 2y = 0
4(-3y + 7) – 2y = 0

3y + x = 7
4x – 2y = 0
the other variable.
4x – 2y = 0
4x – 2(2) = 0
4x – 4 = 0
4x = 4
x = 1
-12y + 28 – 2y = 0
-14y + 28 = 0
-14y = -28
y = 2

3y + x = 7
4x – 2y = 0
Step 5: Check your
solution.
(1, 2)
3(2) + (1) = 7
4(1) – 2(2) = 0
When is solving systems by substitution easier
to do than graphing?
When only one of the equations has a variable
already isolated (like in example #1).

If you solved the first equation for x, what
would be substituted into the bottom equation.
2x + 4y = 4
3x + 2y = 22
1. -4y + 4
2. -2y + 2
3. -2x + 4
4. -2y+ 22

x = 3 – y
x + y = 7
for one variable.
Step 2: Substitute
The first equation is
already solved for x!
x + y = 7
(3 – y) + y = 7
3 = 7
The variables were eliminated!!
This is a special case.
Does 3 = 7? FALSE!
When the result is FALSE, the answer is NO SOLUTIONS.

2x + y = 4
4x + 2y = 8
for one variable.
Step 2: Substitute
The first equation is
easiest to solved for y!
y = -2x + 4
4x + 2y = 8
4x + 2(-2x + 4) = 8
4x – 4x + 8 = 8
8 = 8
This is also a special case.
Does 8 = 8? TRUE!
When the result is TRUE, the answer is INFINITELY MANY SOLUTIONS.

Solving Systems by Graphing and Substitution

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