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You want to look at the questions like fractions and separate them out.
#1 is the same as (2x^2y/2xy)-(16xy^2/2xy)+(12xy/2xy)
#2 becomes (-15xy/5x)+(10x/5x)
You are able to do this since you are dividing by only one term (2xy or 5x). It won't work if you have more than one term. Now if you look at the fractions, look at the numbers, the x's, and the y's separately and reduce what you can like you did with fractions back when you were in elementary school.
#1 For the first fraction, you have a 2/2 in the numbers, so if you remember, that becomes one, or in our case it actually cancels since we have more going on. You have an x^2 on top and an x on the bottom, that leaves you with an x on top. And you have a y/y, so again that's one, or just cancels. So in the end, that first fraction becomes just x.
I'll let you try the rest of #1 and #2 and maybe you'll get what Beckden said for the answer.
= `(2xy(x-8y+6))/(2xy) = x-8y+6`
2. `(-15xy+10x)/(5x) = (-15xy)/(5x) + (10x)/(5x) = -3y+2`
Answers are (1) x-8y+6 and (2) -3y+2
First, you need to use brackets at both numerators, otherwise, the numerator of the 1st expression is just the term 12xy and the numerator of the 12nd expression is just the term 10x.
You have to simplify these two expressions.
We'll begin with
(2x^2*y -16x*y^2 + 12xy) / 2xy
We'll write each term from numerator over the denominator:
2x^2*y -16x*y^2 + 12xy / 2xy = 2x^2*y / 2xy - 16x*y^2 / 2xy + 12xy/ 2xy
We'll use the property of division of two exponentials that have matching bases:
x^a/x^b = x^(a-b)
2x^2*y / 2xy = x^(2-1)*y^(1-1) = x*y^0 = x*1 = x
16x*y^2 / 2xy = (16/2)*x^(1-1)*y^(2-1)
16x*y^2 / 2xy = 8*x^0*y = 8y
12xy/ 2xy = (12/2)*x^(1-1)*y^(1-1) = 6
(2x^2*y -16x*y^2 + 12xy )/ 2xy = x - 8y + 6
We'll simplify the 2nd expression:
(-15xy + 10x) / 5x = -15xy/5x + 10x/5x
-15xy/5x = -(15/5)x^(1-1)*y = -3y
10x/5x = (10/5)*x^(1-1) = 2
(-15xy + 10x) / 5x = -3y + 2
The results of simplifying both expressions are: 1) (2x^2*y -16x*y^2 + 12xy )/ 2xy = x - 8y + 6 and 2) (-15xy + 10x) / 5x = -3y + 2.
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