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1 Warm Up Simplify each expression. 1. 2y 4x 2(4y 2x) 2. 5(x y) + 2x + 5y Write the least common multiple and and and 82...
6-3 Solving Systems by Elimination Warm Up Simplify each expression. 1. 2y – 4x – 2(4y – 2x)
2. 5(x – y) + 2x + 5y
Write the least common multiple. 3. 3 and 6 5. 6 and 8 Holt Algebra 1
4. 4 and 10
6-3 Solving Systems by Elimination
Objectives Solve systems of linear equations in two variables by elimination. Compare and choose an appropriate method for solving systems of linear equations.
Holt Algebra 1
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.
Holt Algebra 1
6-3 Solving Systems by Elimination
Since –2y and 2y have opposite coefficients, the yterm is eliminated. The result is one equation that has only one variable: 6x = –18. When you use the elimination method to solve a system of linear equations, align all like terms in the equations. Then determine whether any like terms can be eliminated because they have opposite coefficients. 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
Later in this lesson you will learn how to multiply one or more equations by a number in order to produce opposites that can be eliminated.
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6-3 Solving Systems by Elimination Example 1: Elimination Using Addition Solve
3x – 4y = 10 by elimination. x + 4y = –2
Step 1 Step 2
3x – 4y = 10 x + 4y = –2 4x + 0 = 8 4x = 8 4x = 8 4 4 x=2
Holt Algebra 1
Write the system so that like terms are aligned. Add the equations to eliminate the y-terms. Simplify and solve for x. Divide both sides by 4.
6-3 Solving Systems by Elimination Example 1 Continued Step 3 x + 4y = –2 2 + 4y = –2 –2 –2 4y = –4 4y –4 4 4 y = –1 Step 4 (2, –1)
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Write one of the original equations. Substitute 2 for x. Subtract 2 from both sides. Divide both sides by 4. Write the solution as an ordered pair.
6-3 Solving Systems by Elimination Check It Out! Example 1 Solve
y + 3x = –2 by elimination. 2y – 3x = 14
y + 3x = –2 2y – 3x = 14 Step 2 3y + 0 = 12 3y = 12
Step 1
Write the system so that like terms are aligned. Add the equations to eliminate the x-terms. Simplify and solve for y. Divide both sides by 3.
y=4 Holt Algebra 1
6-3 Solving Systems by Elimination Check It Out! Example 1 Continued Step 3 y + 3x = –2 4 + 3x = –2 –4 –4 3x = –6 3x = –6 3 3 x = –2 Step 4 (–2, 4)
Holt Algebra 1
Write one of the original equations. Substitute 4 for y. Subtract 4 from both sides. Divide both sides by 3.
Write the solution as an ordered pair.
6-3 Solving Systems by Elimination 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
6-3 Solving Systems by Elimination Example 2: Elimination Using Subtraction Solve
2x + y = –5 by elimination. 2x – 5y = 13
2x + y = –5 –(2x – 5y = 13) 2x + y = –5 –2x + 5y = –13 Step 2 0 + 6y = –18 6y = –18 y = –3
Step 1
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Add the opposite of each term in the second equation. Eliminate the x term. Simplify and solve for y.
6-3 Solving Systems by Elimination Example 2 Continued Step 3 2x + y = –5 2x + (–3) = –5 2x – 3 = –5 +3 +3
Write one of the original equations. Substitute –3 for y.
2x
Simplify and solve for x.
= –2
x = –1 Step 4 (–1, –3)
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Add 3 to both sides.
Write the solution as an ordered pair.
6-3 Solving Systems by Elimination
Remember! Remember to check by substituting your answer into both original equations.
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6-3 Solving Systems by Elimination Check It Out! Example 2 Solve Step 1
Step 2
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3x + 3y = 15 by elimination. –2x + 3y = –5 3x + 3y = 15 –(–2x + 3y = –5) 3x + 3y = 15 + 2x – 3y = +5 5x + 0 = 20 5x = 20 x=4
Add the opposite of each term in the second equation. Eliminate the y term. Simplify and solve for x.
6-3 Solving Systems by Elimination Check It Out! Example 2 Continued Step 3
3x + 3y = 15 3(4) + 3y = 15 12 + 3y = 15 –12 –12 3y = 3 y=1
Step 4
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(4, 1)
Write one of the original equations. Substitute 4 for x. Subtract 12 from both sides. Simplify and solve for y. Write the solution as an ordered pair.
6-3 Solving Systems by Elimination
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 Check It Out! Example 3a Solve the system by elimination. 3x + 2y = 6 –x + y = –2
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6-3 Solving Systems by Elimination Check It Out! Example 3b Solve the system by elimination. 2x + 5y = 26 –3x – 4y = –25
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6-3 Solving Systems by Elimination Check It Out! Example 4 What if…? Sally spent $14.85 to buy 13 flowers. She bought lilies, which cost $1.25 each, and tulips, which cost $0.90 each. How many of each flower did Sally buy?
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6-3 Solving Systems by Elimination
All systems can be solved in more than one way. For some systems, some methods may be better than others.
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6-3 Solving Systems by Elimination
Holt Algebra 1
6-3 Solving Systems by Elimination
6.3 Homework pg. 401: 12-26even
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