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Integer Exponents

Can I have assistance with three problems?

1. (3x^2y-4) (4x^-3y)

2. (15x^-1y^2/5xy^-4)-1

3. 24x^-1y^-2/6x^-4y^3

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4 Answers

Remember these rules of exponents.

aman = am+n, am/an = am-n, and a-n = 1/an.  Also (a/b)-1 = b/a.

Regroup the terms before applying these rules, and you will make everything a little easier.

1. (3x^2y-4) (4x^-3y) = 3*4 x2x-3 y-4y = 12*x-1*y-3 = 12/(xy3).

Try the others using the rules this way.  If you're still having trouble, I or another will be happy to offer further assistance.

Hello Taylor,

These all are exponent problems.

1. (3x2y-4)(4x-3y)

In this problem we are multiplying. First we'll multiply the coefficients and then we'll multiply the variables with bases x and y.


   =12x2-3.y-4+1       (formula xm.xn = xm+n)

   = 12x-1y-3

   = 12/xy3 (answer)  (x-1 = 1/x and y-3 = 1/y3)


2.   (15x-1y2/5xy-4)-1

First we'll take care inside the bracket and solve. Take care of coefficient and then variables.

   = [(15/5).(x-1/x).(y2/y-4)]-1

   = [3(1/x1+1)(y2+4)]-1        (formula used x-1 = 1/x, xm/xn = xm-n

    = [3(1/x2)(y6)]-1

    = (3-1.y6.-1)/x2.-1                (power to power formula (xm)n = xmn)

   = 3-1.y-6/x-2

   = x2/3y6 (answer)


3.  24x-1y-2/6x-4y3

   = (24/6)(x-1/x-4)(y-2/y3)

   = 4(x-1+4)(1/y3+2)    (formula used 1/x-1 = x, x-1 = 1/x, xm/xn = xm-n)

   = 4x3/y5 (answer)

Hope this helps.

Please clarify where the exponents begin and end for each x term.

If problem #1 looks like this: (3x2y-4)(4x-3y) then here's the explanation.

Because we have two terms in the first parenthese, we need to distribute the second parenthese to each of those terms. Thus we have:


Now to simpligy we multiply the constants with constants, x's with x's and y's with y's.

12x2*x-3*y*y-16x-3y. The exponents for the x's add together as per exponential laws (xa*xb=xa+b). Same for the y's:


The x's will need to be brough down because they have a negative exponent (x-a=1/xa).

12y2/x-16y/x3 And lastly we can make a common denominator by multiplying the left term by x2 on both top and bottom and combine the two terms:

(12y2x2-16y)/x3 And that's our simplest answer.

Hope this helped!

1. (3x2y-4)(4x-3y) = (3x4)(x(2-3))(y(-4+1)) = 12x-1y-3

2. (15x-1y2 / 5x y-4)-1 = ( (15/5)(x(-1-1))(y(2+4)) )-1 = ( 3 x-2y6 )-1 = (1/3) x(-2x-1)y(6x-1) = (1/3) x2y-6

3. (24x-1y-2)/(6x-4y3) = (24/6)(x(-1+4))(y(-2-3)) = 4x3y-5

you can use general formula as below

1/xa = x-a , xa time xb = x(a+b) , (xa)b = xab