Solving Systems of Equations by Elimination: 6.3 & 6.4

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Presentation transcript:

Solving Systems of Equations by Elimination: 6.3 & 6.4 Algebra 1: November 3, 2015

Can you and your partner notice anything different about the following system compared to the problems we have been working on in the previous two sections?

Solving by Elimination Align the problem: (vertically) Find opposites. If there are none, multiply to get opposites. Eliminate the variable Solve Solve for the second variable

Use elimination to solve the system of equations Use elimination to solve the system of equations. –3x + 4y = 12 3x – 6y = 18 Since the coefficients of the x-terms, –3 and 3, are additive inverses, you can eliminate the x-terms by adding the equations. Write the equations in column form and add. The x variable is eliminated. Divide each side by –2. y = –15 Simplify. Plug in to solve for x, x = - 24

Use elimination to solve the system of equations Use elimination to solve the system of equations. 4x + 2y = 28 4x – 3y = 18 There are no opposites, but two coefficients are the same number. You can subtract the second equation from the first. REMEMBER to subtract ALL parts!!!

You try a few!

What happens where there are no opposites and we can not subtract? 2x + y = 23 3x + 2y = 37

Multiplying one or both equations to achieve opposites x + 7y = 12 3x – 5y = 10 4x + 2y = -14 5x + 3y = -17

Your Turn!

Four times one number minus three times another number is 12 Four times one number minus three times another number is 12. Two times the first number added to three times the second number is 6. Write a system of linear equations and then use elimination to solve it and find the numbers.

The student council sold pennants and pom-poms. The pom-poms cost $0 The student council sold pennants and pom-poms. The pom-poms cost $0.75 more than the pennants. Two pennants and two pom-poms cost $6.50. What are the prices for the pennants and the pom-poms?