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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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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I don't understand how to use the properties to alter the equation, nothing too big I hope :)
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)
Just find a multiplication in your equation, and change the order of the factors.

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I'm gonna give it a go haha. 3x^2+X (2-x)(X-5)x
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.
Remember to write everything else exactly as i was including the equal sign.
I see what you mean! So it's (5-X)(2-X) correct?
i flipped those on accident, lol. (X-5)(X-2)
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
Since all that changed was the order oft he multiplication, that is an application of the commutative property. That takes care of part A.
Part A. 3x^2+x= (X-5)(X-2)x Ok so far?
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?
So (3x^2)+x= (X-5)(X-2)? I'm not sure really.
Wait. For part A., you are correct just above, but don't forget the x at the end of (x-5)(x-2)x
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.
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.
So would I just change the order of (X-5)(X-2) to (5-x)(2-X)X?
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.
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)
Finally we need to do part C. Now we start with the result of part B. 3x^2+x= x(x - 5)(x - 2)
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)
That is it. This whole problem is just an exercise in using the commutative, associative, and distributive properties.
Sorry, but gtg.
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!
yw

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