Could you please help explain these two algebraic problems? Divide. 1. 2x^2 y (only the x is squared) -16xy^2 + 12xy / 2xy 2. -15xy + 10x / 5x

Asked on by anya4one

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jellybean1977's profile pic

jellybean1977 | High School Teacher | (Level 1) Adjunct Educator

Posted on

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.


Good luck!

beckden's profile pic

beckden | High School Teacher | (Level 1) Educator

Posted on

1. `(2x^2y-16xy^2+12xy)/(2xy)=((2xy)(x)-(2xy)(8y)+(2xy)(6))/(2xy)`
= `(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

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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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