1. Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
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
0 + 12y = 12 Addition Property of Equality
y = 1 Solve for y.
Step 2: Solve for the eliminated variable x using either original equation.
2x + 3y = 11 Choose the first equation.
2x + 3(1) = 11 Substitute 1 for y.
2x + 3 = 11 Solve for x.
2x = 8
x = 4
7-3

2. Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
(continued)
Since x = 4 and y = 1, the solution is (4, 1).
Check: See if (4, 1) makes true the equation not used in Step 2.
–2(4) + 9(1) Substitute 4 for x and 1 for y into the second equation. –
1 = 1
7-3

3. Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
On a special day, tickets for a minor league baseball game were $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
Relate: total number at the game total amount collected
Write: a s = a s = 3067
Solve by elimination.
7-3

4. Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
(continued)
Step 1: Eliminate one variable.
a + s = 1139
5a + s = 3067
–4a = –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.
s = 1139 Substitute 482 for a.
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 , or The money collected was $5(482), or $2410, plus $1(657), or $657, which is $3067. The solution is correct.
7-3

5. Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
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
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
Step 2: Solve for x.
–12x = 48
x = 4
7-3

6. Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
(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.
y = –6 Solve for y.
6y = –18
y = –3
The solution is (4, –3).
7-3

7. Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
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.
Relate: total number of cans total amount of sales
Write: p n = p n = 1614
7-3

8. Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
(continued)
Step 1: Eliminate one variable.
Start with the given
system.
p + n = 240
5p + 8n = 1614
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
Step 2: Solve for n.
–3n = –414
n = 138
7-3

9. Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
(continued)
Step 3: Solve for the eliminated variable using either of the original equations.
p + n = 240 Choose the first equation.
p = 240 Substitute 138 for n.
p = 102 Solve for p.
The band sold 102 cans of popcorn and 138 cans of mixed nuts.
7-3

10. Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
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
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 y = 50
15x y = 30
y = 20
Step 2: Solve for y.
4y = 20
y = 5
7-3

11. Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
(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 = 10
3x = –15
x = –5
The solution is (–5, 5).
7-3

12. Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
Solve using elimination.
1. 3x – 4y = m + 3n = 22
2x + 4y = m + 6n = 34
3. –6x + 5y = p + 5q = 2
3x + 4y = p – 9q = 17
(3, 0.5)
(2, 4)
(1, 2)
(1, 1)
7-3