Download 6-3 Solving Systems by Elimination 6

12/4/2014
6-3 Solving Systems by Elimination
• Correct 6-2
• L to J
• Lesson 6-3
– Solving systems by
Elimination
• Assignment 6-3
Upcoming:
HWQ #13 - Fri. 12/5
6-3 Solving Systems by Elimination
Another method for solving systems of
equations is elimination. Like substitution, the
goal of elimination is to get one equation that
has only one variable. To do this by elimination,
you add the two equations in the system
together.
Remember that an equation stays balanced
if you add equal amounts to both sides. So,
if 5x + 2y = 1, you can add 5x + 2y to one
side of an equation and 1 to the other side
and the balance is maintained.
Quiz: 6-1 to 6-4 - Tue. 12/9
Holt Algebra 1
Holt Algebra 1
6-3 Solving Systems by Elimination
Solving Systems of Equations by
Elimination
Step 1
Write the system so that like
terms are aligned.
Step 2
Eliminate one of the variables and
solve for the other variable.
Step 3
Substitute the value of the variable
into one of the original equations
and solve for the other variable.
Step 4
Write the answers from Steps 2 and 3
as an ordered pair, (x, y), and check.
Holt Algebra 1
6-3 Solving Systems by Elimination
Example 1: Elimination Using Addition
Solve
3x – 4y = 10
by elimination.
x + 4y = –2
Holt Algebra 1
6-3 Solving Systems by Elimination
6-3 Solving Systems by Elimination
Check It Out! Example 1
Solve
Holt Algebra 1
y + 3x = –2
by elimination.
2y – 3x = 14
When two equations each contain
the same term, you can subtract
one equation from the other to
solve the system. To subtract an
equation add the opposite of each
term.
Holt Algebra 1
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12/4/2014
6-3 Solving Systems by Elimination
6-3 Solving Systems by Elimination
Example 2: Elimination Using Subtraction
Solve
2x + y = –5 by elimination.
2x – 5y = 13
Holt Algebra 1
Remember!
Remember to check by substituting your answer
into both original equations.
Holt Algebra 1
6-3 Solving Systems by Elimination
6-3 Solving Systems by Elimination
Check It Out! Example 2
Solve
3x + 3y = 15
by elimination.
–2x + 3y = –5
Holt Algebra 1
6-3 Solving Systems by Elimination
Example 3A: Elimination Using Multiplication First
Solve the system by elimination.
x + 2y = 11
–3x + y = –5
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In some cases, you will first need to
multiply one or both of the equations by
a number so that one variable has
opposite coefficients. This will be the
new Step 1.
Holt Algebra 1
6-3 Solving Systems by Elimination
Example 3B: Elimination Using Multiplication First
Solve the system by elimination.
–5x + 2y = 32
2x + 3y = 10
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12/4/2014
6-3 Solving Systems by Elimination
6-3 Solving Systems by Elimination
Example 4: Application
Check It Out! Example 3a
Solve the system by elimination.
3x + 2y = 6
–x + y = –2
Paige has $7.75 to buy 12 sheets of felt and
card stock for her scrapbook. The felt costs
$0.50 per sheet, and the card stock costs
$0.75 per sheet. How many sheets of each
can Paige buy?
Write a system. Use f for the number of felt
sheets and c for the number of card stock sheets.
0.50f + 0.75c = 7.75
f + c = 12
Holt Algebra 1
6-3 Solving Systems by Elimination
The cost of felt and card
stock totals $7.75.
The total number of sheets
is 12.
Holt Algebra 1
6-3 Solving Systems by Elimination
Homework:
Pages:401-403
11-19 odd,
22-32 even,
33-34,
41-47 odd
Holt Algebra 1
Holt Algebra 1
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