Download Lesson 3.2, 3.3

3.2 Solving by Substitution and
Elimination
3.3 Application
Solving by Substitution
1) Use one equation to get either x or y by itself
2) Substitute what we just found in step 1 to
the other equation
3) Solve for one unknown
4) Substitute the answer from step 3 to either
of the original equation to solve for the
other unknown
5) Check your answer
x+y=4
x=y+2
Step 1) x is already by itself x = y + 2
Step 2) Substitute to the first equation
x +y=4
(y + 2) + y = 4
Step 3)
2y + 2 = 4
2y
=4–2=2
y
=1
Step 4) x = y + 2
x=1+2=3
Solution: (3, 1)
2x + y = 6
3x + 4y = 4
Step 1) get y by itself 2x + y = 6 so y = -2x + 6
Step 2) Substitute to the second equation
3x + 4y = 4
3x + 4(-2x + 6) = 4
Step 3) 3x – 8x + 24 = 4
-5x + 24
=4
-5x
= -20
x
=4
Step 4) 2x + y = 6
2(4)+y = 6 so 8 + y = 6, y = -2
Solution: (4, -2)
More practice: (Make sure you check your answers)
1) 3x – 4y = 14
5x + y = 8
Answer: (2,-2)
2) 2x – 3y = 0
-4x + 3y = -1
Get x by itself: 2x – 3y = 0 so x = (3/2) y
Substitute into: -4x
+ 3y = -1
-4(3/2)y + 3y = -1
-6y + 3y = -1
- 3y = -1 so y = 1/3
Solve for x: Use the first equation 2x – 3y
=0
2x – 3(1/3) = 0
2x
–1
=0
x
=½
Answer: (1/2 , 1/3)
Solve by Elimination
1) Multiply some numbers to either or both
equations to get 2 opposite terms (For
example: 2x and -2x)
2) Add equations to eliminate one variable
3) Solve for 1 unknown
4) Substitute the answer from step 3 to either of
the original equation to solve for the other
unknown
5) Check your answer
• Solve by elimination
2x – 3y = 0
-4x + 3y = -1
Ignore step 1 since we already have 2 opposite terms -3y and 3y
Step 2 &3: Add 2 equations to eliminate y and solve for x
2x – 3y = 0
-4x + 3y = -1
-2x
= -1
x
= 1/2
Step 4: Choose the first equation 2x – 3y = 0
2(1/2) – 3y = 0
1
- 3y = 0
- 3y = -1
y = 1/3
Answer: ( 1/2, 1/3)
Ex2) 2x + 2y = 2
3x – y = 1
Step 1: Multiply 2 to the second equation
2x + 2y = 2
(2) 3x – y = 1
2x + 2y = 2
6x – 2y = 2
8x
= 4 (step 2 and 3)
x
= 1/2
Step 4: Choose 3x – y = 1
3(1/2) – y = 1
3/2 - y = 1
- y = 1 – (3/2) = - ½
y=½
Answer: (1/2, 1/2)
Ex3) 2x + 3y = 8
-3x + 2y = 1
Step 1: Multiply 3 to the 1st equation and 2 to the 2nd equation
(3) 2x + 3y = 8
(2) -3x + 2y = 1
6x + 9y = 24
-6x + 4y = 2
13y = 26 (step 2 and 3)
y =2
Step 4: Choose -3x + 2y = 1
-3x + 2(2) = 1
-3x + 4 = 1
-3x = 1 -4 = -3
x=1
Answer: (1,2)
Practice
Solve by elimination
1) 3x – 4y = 14
5x + y = 8
Answer: (2,-2)
•
3x + 2y = 7
6x + 4y = 14
Answer: Infinite many solutions
• The perimeter of a rectangle is 140 cm. The length is 1
cm more than twice the width. Find the dimensions.
• Let w be the width and L be the length
• Equations: 2w + 2L = 140
L = 1 + 2w
Answer: L = 47 cm
W = 23 cm
• Two angles are supplementary. One angle is 4 degrees
less than three times the other. Find the measures of the
angles.
• Let x and y be two supplementary angles
• Equations: x + y = 180
x = 3y – 4
Answer: 46 degrees and 134 degrees
• Hockey teams receive 2 points when they win and 1
point when they tie. One season, a team won a
championship with 60 points. They won 12 more games
than they tied. How many wins and how many ties did
the team have?
• Let x be the number of wins and y be the number of ties
• Equations: 2x + 1y = 60
x = 12 + y
Answers: 12 ties and 24 wins
• A student makes a $9.50 purchase at the bookstore with a $20 bill.
The store has no bills and gives the change in quarters and fifty-cent
pieces. There are 30 coins in all. How many of each kind are
there?
• Let x be the number of quarters and y be the number of fifty-cent
pieces.
• The store should give back 20 – 9.50 = 10.50 dollars
• Equations: x + y = 30
0.25 x + 0.50 y = 10.50
Solve by sub: y = 30 – x
25x + 50 y = 1050
25x + 50 (30 – x) = 1050
25x + 1500 – 50 x = 1050
- 25x
= - 450
x
= 18, so y = 12
Therefore, there are 18 quarters and 12 of the fifty-cent pieces