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### Transcript

• 1. 6th Grade Math AlgebraJessica Corriere & Robert Richards
• 2. This PowerPoint is asample of the book, 6th Grade Math
• 3. Problem 1) What value of x makes the equation below true? 7 + x = 84 A. 12 B. 77 C. 83 D. 91 Solve simple one-step equations using basic whole- number facts
• 4. How to solve There are 2 basic rules to a problem like this. Isolate the unknown on one side of the equal sign.Whatever action you perform must be performed on both sides of the equal sign. 7 + x = 84 Subtract 7 from both sides to isolate x 7 + x – 7 = 84 – 7 x = 77 The correct answer is Choice B, 77
• 5. Solve simple proportions within contextProblem 2) Guafi selects 15 pieces of fruit. If 3 out of 5 ofthese fruit are bananas, b, how many bananas does Guafiselect?A. 3B. 9C. 13D. 25
• 6. How to solveFacts provided There are 15 total pieces of fruit There are 3 bananas for every 5 pieces of fruit Write a formula that expresses those two statements 15 pieces of fruit, the bananas plus the other fruit, when 3out of every 5 fruit is a banana.G-Zibo Tip – In a word problem, the word “is” means equals in theformula 15 (3/5) = b
• 7. How to solveThere are 2 basic rules to a problem like this.Isolate the unknown on one side of the equal sign. In this case it already is.Whatever action you perform must be performed on both sides of the equal sign. 15 (3/5) = bSimplify the left sideMultiply 3/5 times 153 x 15 = b 545 = b59=b Choice B is the correct answer
• 8. Translate two-step verbal sentences into algebraic equationsProblem 3) What algebraic equation represents “three timesthe difference of a number, x, and nine equals fifteen”?A. 9x -3 = 15B. 3(9) – x = 15C. 9 – 3x = 15D. 3(x – 9) = 15
• 9. How to solve The statement gives 3 times the difference of x and 9 3(x – 9) equals 15the word “equals” determines where the equal sign goes,15 on one side,everything else on the other 3(x -9) = 15Choice D is the correct answer
• 10. Click book covers to orderThe book contains the same type questions found on the state test with full explanations on how to solve each of them Available in paperback and eBook eBook – click eBook – click to view to view Paperback – click to Paperback – click to view view