Homework 12/15/2015
Solving Systems of linear Equations packet
Page 1, 2, and 3
Note: I am not available after school =(
Systems of Linear Equations
A solution to a system of equations is an ordered
pair that satisfy all the equations in the system.
A system of linear equations can have:
• 1. Exactly one solution
• 2. No solutions
• 3. Infinitely many solutions
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Consistent
Inconsistent
One solution
No solution
Lines intersect
Lines are parallel
Dependent
Infinite number of
solutions
Coincide-Same
4
line
There are four ways to solve systems of linear
equations:
1. By graphing
2. By substitution
3. By elimination
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 Solving Systems by Graphing:
 When solving a system by graphing:
1. Change the equations into y = mx + b so you can
see the slope (m) and the y-intercept (b)
2. Graph both equations
3. Find the intersection point (break-even point)
and that will determine the solution or solutions
to the system of equations.
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
Three possible solutions to a linear system in two
variables:
One solution: coordinates of a point
No solutions: inconsistent case
Infinitely many solutions: dependent case
2x – y = 2
x + y = -2
2x – y = 2
-y = -2x + 2
y = 2x – 2
x + y = -2
y = -x - 2
Different slope, different intercept!
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3x + 2y = 3
3x + 2y = -4
3x + 2y = 3
2y = -3x + 3
y = -3/2 x + 3/2
3x + 2y = -4
2y = -3x -4
y = -3/2 x - 2
Same slope, different intercept!!
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x – y = -3
2x – 2y = -6
x – y = -3
-y = -x – 3
y =x+3
2x – 2y = -6
-2y = -2x – 6
y=x+3
Same slope, same intercept!
Same equation!!
Determine Without Graphing:
Once the equations are in slope-intercept form,
compare the slopes and intercepts.
One solution – the lines will have different slopes.
No solution – the lines will have the same slope,
but different intercepts.
Infinitely many solutions – the lines will have the
same slope and the same intercept.
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Determine Without Graphing:
Given the following lines, determine what type of
solution exists, without graphing.
Equation 1:
3x = 6y + 5
Equation 2:
y = (1/2)x – 3
Writing each in slope-intercept form (solve for y)
Equation 1:
y = (1/2)x – 5/6
Equation 2:
y = (1/2)x – 3
Since the lines have the same slope but different yintercepts, there is no solution to the system of
equations. The lines are parallel.
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Substitution Method:
Procedure for Substitution Method
1. Solve one of the equations for one of the variables.
2. Substitute the expression found in step 1 into the
other equation.
3. Now solve for the remaining variable.
4. Substitute the value from step 2 into the equation
 written in step 1, and solve for the remaining
variable.
Substitution Method:
1. Solve the following system of equations by
substitution.
y  x3
x  y  5
Step 2:Substitute x+3 into
2nd equation and solve.
x  ( x  3)  5
2x  3  5
2x  8
x  4
Step 1 is already completed.
Step 3: Substitute –4 into 1st
equation and solve.
y  x3
y  4  3
y  1
The answer: ( -4 , -1)
1) Solve the system using substitution
x+y=5
y=3+x
Step 1: Solve an
equation for one
variable.
Step 2: Substitute
Step 3: Solve the
equation.
The second equation is
already solved for y!
x+y=5
x + (3 + x) = 5
2x + 3 = 5
2x = 2
x=1
1) Solve the system using substitution
x+y=5
y=3+x
Step 4: Plug back in to
find the other
variable.
Step 5: Check your
solution.
x+y=5
(1) + y = 5
y=4
(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?
2) Solve the system using substitution
3y + x = 7
4x – 2y = 0
Step 1: Solve an
equation for one
variable.
It is easiest to solve the
first equation for x.
3y + x = 7
-3y
-3y
x = -3y + 7
Step 2: Substitute
4x – 2y = 0
4(-3y + 7) – 2y = 0
2) Solve the system using substitution
3y + x = 7
4x – 2y = 0
Step 3: Solve the
equation.
Step 4: Plug back in to
find the other
variable.
-12y + 28 – 2y = 0
-14y + 28 = 0
-14y = -28
y=2
4x – 2y = 0
4x – 2(2) = 0
4x – 4 = 0
4x = 4
x=1
2) Solve the system using substitution
3y + x = 7
4x – 2y = 0
Step 5: Check your
solution.
(1, 2)
3(2) + (1) = 7
4(1) – 2(2) = 0
Solving Systems of Equations
using Elimination
Steps:
1. Place both equations in Standard Form, Ax + By = C.
2. Determine which variable to eliminate with Addition
or Subtraction.
3. Solve for the remaining variable.
4. Go back and use the variable found in step 3 to find
the second variable.
5. Check the solution in both equations of the system.
EXAMPLE #1:
5x + 3y = 11
5x = 2y + 1
STEP1: Write both equations in Ax + By = C
form.
5x + 3y =1
5x - 2y =1
STEP 2:
Use subtraction to eliminate 5x.
5x + 3y =11
5x + 3y = 11
-(5x - 2y =1)
-5x + 2y = -1
Note: the (-) is distributed.
STEP 3:
Solve for the variable.
5x + 3y =11
-5x + 2y = -1
5y =10
y=2
5x + 3y = 11
5x = 2y + 1
STEP 4:
Solve for the other variable by substituting
into either equation.
5x + 3y =11
5x + 3(2) =11
5x + 6 =11
5x = 5
x=1
The solution to the system is (1,2).
5x + 3y= 11
5x = 2y + 1
Step 5:
Check the solution in both equations.
The solution to the system is (1,2).
5x + 3y = 11
5(1) + 3(2) =11
5 + 6 =11
11=11
5x = 2y + 1
5(1) = 2(2) + 1
5=4+1
5=5
Example #2:
x + y = 10
5x – y = 2
Step 1: The equations are already in standard
form:
x + y = 10
5x – y = 2
Step 2: Adding the equations will eliminate y.
x + y = 10
x + y = 10
+(5x – y = 2)
+5x – y = +2
Step 3:
Solve for the variable.
x + y = 10
+5x – y = +2
6x = 12
x=2
x + y = 10
5x – y = 2
Step 4:
Solve for the other variable by
substituting into either equation.
x + y = 10
2 + y = 10
y=8
Solution to the system is (2,8).
x + y = 10
5x – y = 2
Step 5:
Check the solution in both equations.
Solution to the system is (2,8).
x + y =10
2 + 8 =10
10=10
5x – y =2
5(2) - (8) =2
10 – 8 =2
2=2
NOW solve these using elimination:
1.
2.
2x + 4y =1
x - 4y =5
2x – y =6
x+y=3
Using Elimination to Solve a
Word Problem:
Two angles are supplementary. The
measure of one angle is 10 degrees
more than three times the other.
Find the measure of each angle.
Using Elimination to Solve a
Word Problem:
Two angles are supplementary. The
measure of one angle is 10 more
than three times the other. Find the
measure of each angle.
x = degree measure of angle #1
y = degree measure of angle #2
Therefore x + y = 180
Using Elimination to Solve a
Word Problem:
Two angles are supplementary. The
measure of one angle is 10 more
than three times the other. Find the
measure of each angle.
x + y = 180
x =10 + 3y
Using Elimination to Solve a
Word Problem:
Solve
x + y = 180
x =10 + 3y
x + y = 180
-(x - 3y = 10)
4y =170
y = 42.5
x + 42.5 = 180
x = 180 - 42.5
x = 137.5
(137.5, 42.5)
Using Elimination to Solve a
Word Problem:
The sum of two numbers is 70
and their difference is 24. Find
the two numbers.
Using Elimination to Solve a
Word problem:
The sum of two numbers is 70
and their difference is 24. Find
the two numbers.
x = first number
y = second number
Therefore, x + y = 70
Using Elimination to Solve a
Word Problem:
The sum of two numbers is 70
and their difference is 24. Find
the two numbers.
x + y = 70
x – y = 24
Using Elimination to Solve a
Word Problem:
x + y =70
x - y = 24
2x = 94
x = 47
47 + y = 70
y = 70 – 47
y = 23
(47, 23)
Now you Try to Solve These
Problems Using Elimination.
Solve
1. Find two numbers whose sum is
18 and whose difference is 22.
2. The sum of two numbers is 128
and their difference is 114. Find
the numbers.