1. 6-3 Solving Systems Using Elimination (Combination)
Hubarth
Algebra

2. Ex 1 Adding Equations
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
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) Substitute 4 for x and 1 for y into the
second equation. –
1 = 1

3. Ex 2 Application
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
Relate: total number at the game total amount collected
Write: a s = a s = 3067
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.

4. Ex 3 Multiplying One Equation
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
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).

5. Ex 4 Multiplying Both Equations
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
3x + 5y = 10 Use the first equation.
3x + 5(5) = 10 Substitute 5 for y.
3x = 10
3x = –15
x = –5
Step 3: Solve for the eliminated variable x using either of the original
equations.
The solution is (–5, 5).

6. Practice
1. Solve by Elimination
x + y = 10
x – y = 8
b. -2x + 15y = -32
7x – 5y = 17
c. 15x + 3y = 9
10x + 7y = -4
(9, 1)
(1, -2)
(1, -2)
2. Your class sells a total of 64 tickets to a play. A student tickets cost $1, and an adult ticket
Costs $2.50. Your class collects $109 in total ticket sales. How many adult tickets did you sell?
How many student tickets did you sell?
30 adults, 34 students