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GenericNoodles

  • one year ago

Algebra 1 question! Consider the equation 3x^2+x= (X-2)(X-5)x A. Use the commutative property to create an equation with the same solution set. B. Using the result from part A, use the associative property to create an equation with the same solution set. C. using the results from part b, use the distributive property to create an equation with the same solution set.

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  1. GenericNoodles
    • one year ago
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    I don't understand how to use the properties to alter the equation, nothing too big I hope :)

  2. mathstudent55
    • one year ago
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    It's easier that you think. The commutative property is that you can change the order of the factors in a multiplication without changing the answer. Examples: ab = ba 5 * 4 = 4 * 5 (x + 1)(x - 3) = (x - 3)(x + 1)

  3. mathstudent55
    • one year ago
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    Just find a multiplication in your equation, and change the order of the factors.

  4. GenericNoodles
    • one year ago
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    I'm gonna give it a go haha. 3x^2+X (2-x)(X-5)x

  5. mathstudent55
    • one year ago
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    No. When you changed from x - 2 to 2 - x, you actually changed the factors. x - 2 is not the same as 2 - x. Keep x - 2 exactly as it is. Just change the order of the factors x - 2 and x - 5.

  6. mathstudent55
    • one year ago
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    Remember to write everything else exactly as i was including the equal sign.

  7. GenericNoodles
    • one year ago
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    I see what you mean! So it's (5-X)(2-X) correct?

  8. GenericNoodles
    • one year ago
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    i flipped those on accident, lol. (X-5)(X-2)

  9. mathstudent55
    • one year ago
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    It is by accident, but DO NOT CHANGE the content of each factor. Keep x - 2 and x - 5 as they are. Original equation: 3x^2+x= (X-2)(X-5)x This is what I was going for: 3x^2+x= (X-5)(X-2)x See how x - 2 remains as x - 2 and x - 5 remains as x - 5? What did change is that the right side was originally (X-2)(X-5)x, and now it's (X-5)(X-2)x

  10. mathstudent55
    • one year ago
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    Since all that changed was the order oft he multiplication, that is an application of the commutative property. That takes care of part A.

  11. mathstudent55
    • one year ago
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    Part A. 3x^2+x= (X-5)(X-2)x Ok so far?

  12. GenericNoodles
    • one year ago
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    I'm sorry for the confusion, it's been a long day! 3x^2+X=(X-5)(X-2) Associative is putting them into groups yes?

  13. GenericNoodles
    • one year ago
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    So (3x^2)+x= (X-5)(X-2)? I'm not sure really.

  14. mathstudent55
    • one year ago
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    Wait. For part A., you are correct just above, but don't forget the x at the end of (x-5)(x-2)x

  15. mathstudent55
    • one year ago
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    Now for part B, you need the associative property. The associative property states that you can change the grouping in a multiplication without changing the answer.

  16. mathstudent55
    • one year ago
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    For part B, you start with the answer to part A. Let's start with this: 3x^2+x= (x - 5)(x - 2)x Now we use the associative property.

  17. GenericNoodles
    • one year ago
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    So would I just change the order of (X-5)(X-2) to (5-x)(2-X)X?

  18. mathstudent55
    • one year ago
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    If you look on the right side, according to the order of operations (you may know it as PEMDAS), we do multiplications in the order they appear from left to right.

  19. mathstudent55
    • one year ago
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    That means you'd do first (x - 5) times (x - 2), then multiply the result by x. That means x - 5 is grouped with x - 2. Then you multiply by x. To change the grouping, just write the x first. Then you'll be grouping the x with x - 5. That is a change of grouping. Part B. 3x^2+x= x(x - 5)(x - 2)

  20. mathstudent55
    • one year ago
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    Finally we need to do part C. Now we start with the result of part B. 3x^2+x= x(x - 5)(x - 2)

  21. mathstudent55
    • one year ago
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    The distributive property is: a(b + c) = ab + ac or in reverse ab + ac = a(b + c) We use the reverse on the left side to factor out an x. Part C. x(3x + 1)= x(x - 5)(x - 2)

  22. mathstudent55
    • one year ago
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    That is it. This whole problem is just an exercise in using the commutative, associative, and distributive properties.

  23. mathstudent55
    • one year ago
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    Sorry, but gtg.

  24. GenericNoodles
    • one year ago
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    I really appreciate you explaining this out to me! I didn't catch it in class but it's very clear to me now. Thank you!

  25. mathstudent55
    • one year ago
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    yw

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