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

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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Remember these rules of exponents.

a^{m}a^{n} = a^{m+n}, a^{m}/a^{n} = a^{m-n}, and a^{-n
}= 1/a^{n}. 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 x^{2}x^{-3} y^{-4}y = 12*x^{-1}*y^{-3} = 12/(xy^{3}).

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. (3x^{2}y^{-4})(4x^{-3}y)

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

3.4.x^{2}.x^{-3}.y^{-4}.y

=12x^{2-3}.y^{-4+1} (formula x^{m}.x^{n} = x^{m+n})

= 12x^{-1}y^{-3}

= **12/xy3 (answer)** (x^{-1} = 1/x and y^{-3} = 1/y^{3})

2. (15x^{-1}y^{2}/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).(y^{2}/y^{-4})]^{-1}

= [3(1/x^{1+1})(y^{2+4})]^{-1} (formula used x^{-1} = 1/x, x^{m}/x^{n} = x^{m-n})

= [3(1/x^{2})(y^{6})]^{-1}

= (3^{-1}.y^{6.-1})/x^{2.-1 }(power to power formula (x^{m})^{n} = x^{mn})

= 3^{-1}.y^{-6}/x^{-2}

= **x ^{2}/3y^{6} (answer)**

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

= (24/6)(x^{-1}/x^{-4})(y^{-}^{2}/y^{3})

= 4(x^{-1+4})(1/y^{3+2}) (formula used 1/x^{-1} = x, x^{-1} = 1/x, x^{m}/x^{n} = x^{m-n})

= **4x ^{3}/y^{5} (answer)**

Hope this helps.

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

If problem #1 looks like this: (3x^{2}y-4)(4x^{-3}y) 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:

3x^{2}y*4x^{-3}y-4*4x^{-3}y

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

12x^{2}*x^{-3}*y*y-16x^{-3}y. The exponents for the x's add together as per exponential laws (x^{a}*x^{b}=x^{a+b}). Same for the y's:

12x^{-1}y^{2}-16x^{-3}y

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

12y^{2}/x-16y/x^{3} And lastly we can make a common denominator by multiplying the left term by x^{2} on both top and bottom and combine the two terms:

(12y^{2}x^{2}-16y)/x^{3} And that's our simplest answer.

Hope this helped!

1. (3x^{2}y^{-4})(4x^{-3}y) = (3x4)(x^{(2-3)})(y^{(-4+1)}) = 12x^{-1}y^{-3
}

2. (15x^{-1}y^{2} / 5x y^{-4})^{-1} = ( (15/5)(x^{(-1-1)})(y^{(2+4)}) )^{-1
}= ( 3 x^{-2}y^{6} )^{-1} = (1/3) x^{(-2x-1)}y^{(6x-1)} = (1/3) x^{2}y^{-6}

3. (24x^{-1}y-2)/(6x^{-4}y^{3}) = (24/6)(x^{(-1+4)})(y^{(-2-3)}) = 4x^{3}y^{-5}

you can use general formula as below

1/x^{a }= x^{-a} , x^{a} time x^{b} = x^{(a+b)} , (x^{a})^{b} = x^{ab}