Lesson 6-3 Warm-Up
ALGEBRA 1
“Solving Systems Using
Elimination” (6-3)
What is the
elimination method
for solving a system
of equations?
Method 3: Elimination Method: If the coefficients for one of the terms in both
equations has a sum or difference of zero (in other words, the terms are opposites,
like 6x and -6x or d and -d), you can use The Addition or Subtraction Properties of
Equality to eliminate the variable. Then, you can solve for the variable that is not
eliminated and use substitution to find the other variable.
Example: Solve 5x – 6y = -32 and 3x + 6y = 48
Step 1: Add the equations to eliminate the 6y and -6y (opposite terms) and solve
for x.
Step 2: Solve for y by substituting the solution for x into one of the original
equations.
Since x = 2 and y = 7, the solution is (2, 7)
Step 3: Check if the solution makes both equations true statements
5x – 6y = -32
3x + 6y = 48
Start with original equations.
5(2) - 6(7) = – 33
3(2) + 6(7) = 48
y = 7 and x = 2 (substitute)
32 = 32
48 = 48
ALGEBRA 1
Solving Systems Using Elimination
LESSON 6-3
Additional Examples
Solve by elimination.
2x + 3y = 11
–2x + 9y = 1
Step 1: Eliminate x because the sum of the coefficients is 0.
2x + 3y = 11
–2x + 9y =1
Addition Property of Equality
0 + 12y = 12
y=1
Solve for y.
Step 2: Solve for the eliminated variable x using either original equation.
2x + 3y = 11
2x + 3(1) = 11
2x + 3 = 11
2x = 8
x=4
Choose the first equation.
Substitute 1 for y.
Solve for x.
ALGEBRA 1
Solving Systems Using Elimination
LESSON 6-3
Additional Examples
(continued)
Since x = 4 and y = 1, the solution is (4, 1).
Check: See if (4, 1) makes the equation not used in Step 2 true.
–2(4) + 9(1) 1
Substitute 4 for x and 1 for y into the
second equation.
–8 + 9
1
1 = 1
ALGEBRA 1
Solving Systems Using Elimination
LESSON 6-3
Additional Examples
On a special day, tickets for a minor league baseball
game cost $5 for adults and $1 for students. The attendance that
day was 1139, and $3067 was collected. Write and solve a system
of equations to find the number of adults and the number of
students that attended the game.
Define: Let
a
= number of adults
Let
s
= number of students
Words: total number at the game
Equation: a
+ s
= 1139
total amount collected
5 a
+ s = 3067
Since you have an “s” in both equations, solve by subtracting the equations
ALGEBRA 1
Solving Systems Using Elimination
LESSON 6-3
Additional Examples
(continued)
Step 1: Eliminate one variable.
a + s = 1139
(-5a) + (-s) = (-3067) -1(5a + s = 3067) = [-5a + (-s) = -3067]
–4a + 0 = –1928 Subtraction Property of Equality
a
= 482
Solve for a.
Step 2: Solve for the eliminated variable using either of the original
equations.
a + s = 1139
Choose the first equation.
(482) + s = 1139
Substitute 482 for a.
-482
-482
Subtract 482 from both sides.
s = 657
Solve for s.
There were 482 adults and 657 students at the game.
Check: Is the solution reasonable? The total number at the game was
482 + 657, or 1139. The money collected was $5(482), or $2410,
plus $1(657), or $657, which is $3067. The solution is correct.
ALGEBRA 1
“Solving Systems Using
Elimination” (6-3)
How can you
eliminate one of
the variables if
the sum or
difference of one
of the variables
doesn’t equal 0?
Sometimes, you may need to multiply one or both equations in a system
of equations by different nonzero numbers in order to eliminate one of
the variables by adding or subtracting the equations.
Example: Solve 4x + 2y = 14 and 7x - 3y = -8
Step 1: We see that the y coefficients, 2 and 3 can become opposites if
we multiplied them by each other. Therefore, we’ll multiply the equation
with 2y by 3 and the equation with -3y by 2 to create equivalent
equations with opposite terms. Then, we can add the two equivalent
equations together to eliminate one of the equations.
Step 2: Solve for x.
26x = 26
26
26
x=1
Step 1
Divide both sides by 26.
ALGEBRA 1
“Solving Systems Using
Elimination” (6-3)
Step 3: Solve for the eliminated variable, y, by substituting the solution
for x into one of the original equations.
4x + 2y = 14
Given
4(1) + 2y = 14
Substitute (x = 1)
-4
-4
Subtract 4 from both side
2y = 10
Simplify
2
2
Divide both side by 2
y=5
Since x = 1 and y = 5, the solution is (1, 5)
Step 4: Check if the solution makes both equations true statements
4(1) + 2(5) = 14
7(1) – 3(5) = -8
4 + 10 = 14
7 - 15 = -8 y = 7 and x = 2
(substitute)
14 = 14
-8 = -8
ALGEBRA 1
Solving Systems Using Elimination
LESSON 6-3
Additional Examples
Solve by elimination.
3x + 6y = –6
–5x – 2y = –14
Step 1: Eliminate one variable.
Start with the given
system.
3x + 6y = –6
–5x – 2y = –14
Step 2: Solve for x.
–12x = –48
x= 4
To prepare to eliminate
y, multiply the second
equation by 3.
3x + 6y = –6
3(–5x – 2y = –14)
Add the equations to
eliminate y.
3x + 6y = –6
–15x – 6y = –42
–12x + 0 = –48
Divide both sides by -12.
ALGEBRA 1
Solving Systems Using Elimination
LESSON 6-3
Additional Examples
(continued)
Step 3: Solve for the eliminated variable using either of the original
equations.
3x + 6y = –6
Choose the first equation.
3(4) + 6y = –6
Substitute 4 for x.
12 + 6y = –6
Subtract 12 from both sides.
6y = –18
Divide both sides by 6.
y = –3
The solution is (4, –3).
ALGEBRA 1
Solving Systems Using Elimination
LESSON 6-3
Additional Examples
Suppose the band sells cans of popcorn for $5 per can
and cans of mixed nuts for $8 per can. The band sells a total of
240 cans and receives a total of $1614. Find the number of cans
of popcorn and the number of cans of mixed nuts sold.
Define: Let
p
= number of cans of popcorn sold.
Let
n
= number of cans of nuts sold.
Words: total number of cans
total amount of sales
Equations:
5 p
p
+ n
= 240
+ 8 n = 1614
ALGEBRA 1
Solving Systems Using Elimination
LESSON 6-3
Additional Examples
(continued)
Step 1: Eliminate one variable.
Start with the given
system.
p + n = 240
5p + 8n = 1614
Step 2: Solve for n.
–3n = –414
n = 138
To prepare to eliminate
p, multiply the first
equation by 5.
5(p + n = 240)
5p + 8n = 1614
Subtract the equations
to eliminate p.
5p + 5n = 1200
- (5p + 8n = 1614)
0 – 3n = –414
Divide both sides by -3.
ALGEBRA 1
Solving Systems Using Elimination
LESSON 6-3
Additional Examples
(continued)
Step 3: Solve for the eliminated variable using either of the original
equations.
p + n = 240
Choose the first equation.
p + 138 = 240
Substitute 138 for n.
p = 102
Solve for p.
The band sold 102 cans of popcorn and 138 cans of mixed nuts.
ALGEBRA 1
Solving Systems Using Elimination
LESSON 6-3
Additional Examples
Solve by elimination.
3x + 5y = 10
5x + 7y = 10
Step 1: Eliminate one variable.
Start with the given
system.
3x + 5y = 10
5x + 7y = 10
Step 2: Solve for y.
4y = 20
y = 5
To prepare to eliminate
x, multiply one equation
by 5 and the other
equation by 3.
5(3x + 5y = 10)
3(5x + 7y = 10)
Subtract the equations
to eliminate x.
15x + 25y = 50
15x + 21y = 30
0 + 4y = 20
Divide both sides by 4.
ALGEBRA 1
Solving Systems Using Elimination
LESSON 6-3
Additional Examples
(continued)
Step 3: Solve for the eliminated variable x using either of the original
equations.
3x + 5y = 10
Use the first equation.
3x + 5(5) = 10
Substitute 5 for y.
3x + 25 = 10
Subtract 25 from both sides.
3x = –15
x = –5
The solution is (–5, 5).
ALGEBRA 1
Solving Systems Using Elimination
LESSON 6-3
Lesson Quiz
Solve using elimination.
1. 3x – 4y = 7
2x + 4y = 8
(3, 0.5)
3. –6x + 5y = 4
3x + 4y = 11
(1, 2)
2. 5m + 3n = 22
5m + 6n = 34
(2, 4)
4. 7p + 5q = 2
8p – 9q = 17
(1, 1)
ALGEBRA 1