1 / 16

10-3

Solving Equations with Variables on Both Sides. 10-3. Warm Up. Problem of the Day. Lesson Presentation. Pre-Algebra. What are some observations you made about solving equations with variables on both sides, based on last night’s homework?

farrah-velasquez
Download Presentation

10-3

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Solving Equations with Variables on Both Sides 10-3 Warm Up Problem of the Day Lesson Presentation Pre-Algebra

  2. What are some observations you made about solving equations with variables on both sides, based on last night’s homework? • Goal is to isolate the variable – get variables on one side and numbers on the other. • You can go in different order. There are different ways to go about the problem.

  3. –3x 3 = –3 3 6 –3 What can NOT do? Ex A) 4x + 6 = x 4x + 6 = x 4x + 6 = x – 4x– 4x –x -6 – x -6 6 = –3x 3x = –6 –2 = x x = -2

  4. 4b 24 = 4 4 Ex B) 9b – 6 = 5b + 18 9b – 6 = 5b + 18 – 5b– 5b 4b – 6 = 18 + 6+ 6 4b = 24 b = 6

  5. With each step, you are keeping the equation, ajd just simplifying. If your final answer turns out to be not true, there is NO SOLUTION that would work for this equation. Ex.C) 9w + 3 = 5w + 7 + 4w 9w + 3 = 5w + 7 + 4w 9w + 3 = 9w + 7 – 9w– 9w 3 ≠ 7 No solution. There is no number that can be substituted for the variable w to make the equation true.

  6. 7 10 7 10 7 10 3y 5 3y 5 3y 5 y 5 y 5 y 5 3 4 3 4 3 4 20( ) = 20( ) + – y – 20() + 20( ) – 20( )= 20(y) – 20( ) Ex. D + – = y – Multiply by the LCD. 4y + 12y – 15 = 20y – 14 16y – 15 = 20y – 14 Combine like terms.

  7. 4y 4 –1 -1 4 = y = 4 16y – 15 = 20y – 14 – 16y– 16y Subtract 16y from both sides. –15 = 4y – 14 + 14+ 14 Add 14 to both sides. –1 = 4y Divide both sides by 4.

  8. 10z50 = 10 10 Ex.E) 12z – 12 – 4z = 6 – 2z + 32 12z – 12 – 4z = 6 – 2z + 32 8z – 12 = –2z + 38 + 2z+ 2z 10z – 12 = + 38 + 12+12 10z = 50 z = 5

  9. Additional Example 3: Consumer Application Jamie spends the same amount of money each morning. On Sunday, he bought a newspaper for $1.25 and also bought two doughnuts. On Monday, he bought a newspaper for fifty cents and bought five doughnuts. On Tuesday, he spent the same amount of money and bought just doughnuts. How many doughnuts did he buy on Tuesday?

  10. = 3d 3 0.75 3 Additional Example 3 Continued First solve for the price of one doughnut. d = price of one doughnut. 1.25 + 2d = 0.50 + 5d – 2d– 2d 1.25 = 0.50 + 3d – 0.50– 0.50 0.75 = 3d The price of one doughnut is $0.25. 0.25 = d

  11. 1.75 0.25 = 0.25n 0.25 Additional Example 3 Continued Now find the amount of money Jamie spends each morning. Choose one of the original expressions. 1.25 + 2d 1.25 + 2(0.25) = 1.75 Jamie spends $1.75 each morning. Find the number of doughnuts Jamie buys on Tuesday. Let n represent the number of doughnuts. 0.25n = 1.75 Divide both sides by 0.25. n = 7; Jamie bought 7 doughnuts on Tuesday.

More Related