3-2 Solving Systems Algebraically
Hubarth
Algebra II
Ex 1 Solve Systems by Substitution
Solve using substitution.
y = 2x + 2
y = -x + 5
Step 1: Write an equation containing only one variable and solve.
y = -x + 5
y = 2x + 2
–x + 5= 2x + 2
5 = 3x + 2
3 = 3x
1=x
Substitute –x + 5 for y in that equation.
Add x to each side.
Subtract 2 from each side.
Divide each side by 3.
Step 2: Solve for the other variable.
y = 2(1) + 2
y=2+2
y=4
The solution is ( 1, 4)
Substitute 1 for x in either equation.
Simplify.
Ex 2 Using Substitution and the Distributive Property
Solve using substitution.
−𝑥 + 2𝑦 = 8
2𝑦 + 2𝑥 = 2
Step 1: Solve the first equation for x because it has a coefficient of -1.
−𝑥 + 2𝑦 = 8
−𝑥 = −2𝑦 + 8
𝑥 = 2𝑦 − 8
Add x to each side.
Divide all parts by -1
Step 2: Write an equation containing only one variable and solve.
2𝑦 + 2𝑥 = 2
2𝑦 + 2(2y − 8) = 2
2𝑦 + 4𝑦 − 16 = 2
6𝑦 − 16 = 2
Start with the other equation.
Substitute 2𝑦 − 8 for x in that equation.
Use the Distributive Property.
Combine like terms and add16 to each side.
y=3
Divide each side by 6.
Step 3: Solve for y in the other equation.
-x + 2(3) = 8
-x + 6 =8
x = -2
Substitute 3 for y.
Simplify and divide each side by -1.
Subtract 2 from each side.
(-2, 3)
Ex 3 Solve System by Using Substitution
y = (4x – 1)
5 = 6x –(y )
5 = 6x – (4x – 1)
y = 4 (2) – 1
5 = 6x – 4x + 1
y=8–1
5 = 2x +1
y=7
4 = 2x
x=2
(2, 7) is the solution of the system
Ex 4 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) 1
–8 + 9
1
1 = 1
Substitute 4 for x and 1 for y into the
second equation.
Ex 5 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
Write:
a
+ s
= 1139
total amount collected
5 a +1 s = 3067
Step 1: Eliminate one variable.
a + s = 1139
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.
s = 657
Solve for s.
There were 482 adults and 657 students at the game.
Ex 6 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.
12 + 6y = –6
Solve for y.
6y = –18
y = –3
The solution is (4, –3).
Ex 7 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
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
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
3x = –15
x = –5
The solution is (–5, 5).
Practice
1. Solve by Elimination
a. 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. At Renaldi’s Pizza, a soda and two slices of pizza-of-the-day costs $10.25. A soda and
four slices of the pizza-of-the-day costs $18.75. Find the cost of each item?
x = 1.75, y = 4.25
3. x – 5 = y
2x – 3y = 7
(8, 3)
4. x = 4y – 1
3x + 5y = 31
(7, 2)
5. 𝑦 = −2𝑥 + 1
1
5
𝑦 = −2x + 2
(-1, 3)
Practice
1. y = x – 3
x+y=5
(4, 1)
3. x – 5 = y
2x – 3y = 7
(8, 3)
2. 2x + y = 1
x = -2y + 5
(-1, 3)
4. x = 4y – 1
3x + 5y = 31
(7, 2)