Reducing rational expressions?

Question

Here's my question:

x^5y^3z^2 / x^2y^3z^7

In my textbook, it shows how to reduce this expression like this:

x^5-^2 y^3-^3 z^2-^2 / x^2-^2 y^3-^3 z^7-^2

Why is this?? And then it shows that this expression reduces to:

x^3y^0z^0 / x^0y^0z^5 = x^3/z^5

How does it end up being x^3/z^5?
Do the numbers with the exponent of 0 (x^0) end up being omitted or something?

Answer 1

x^5y^3z^2 / x^2y^3z^7
The LCD is x^2y^3z^2
Divide numerator and denominator by the LCD:
x^5/x^2=x^(5-2), and so forth
=x^(5-2)y^(3-3)z^(2-2) / x^2-2)y(3-3)z^(7-2)
=x^3y^0z^0 / x^0y^0z^5
=(x^3)*1*1 / 1*1*(z^5)
=x^3 / z^5

Below is a simpler approach:
=(x^5/x^2)(v^3/y^3)(z^2/z^7)
=x^(5-2)y^(3-3)z^(2-7)
=(x^3)*1*(z^-5)
=x^3/z^5

Answer 2

The reason is x^0 = 1 (anything to the power 0 is 1)
Easiest way to do this division is to subtract the exponent values in the denominator from the value in the numerator for the same base.
We get x^(5-2)y^(3-3)z^(2-7) = x^3y^0z^-5 which is the same as
x^3/y^5

Answer 3

When there is an exponent placed on a same base on both the bottom and the top, you can minus the exponents starting from the top to the bottom.
so, (x^5) / (x^2) = (x^3)
and (y^3) / (y^3) = 1
and (z^2) / (z^7) = 1 / (z^5)
now, put all the back in (multiply), so (x^3) / (z^5)

Answer 4

when dividing variables(x,y,z) you subtract the exponents of the same variable e.g x^5-x^2=x^3
so it can also be done this way x^(5-2)y^(3-3)z^(2-7).......
..................................................
=x^3y^0z^-5
1)remember any variable with zero exponent =1 i.e 2^0 =1, x^0=1 and any variable to the negative exponent can be written as the reciprocal e.g z^-5 is the same as 1/z^5 anthr e.g is say
x^-2=1/x^2

2)In the case of your text bk the just simply subtract the smaller exponent from the larger one

Answer 5

a million. (((2 / 5y) - (3 / 15y)) / 2) / ((4 / 7) + (7 / 15y)) initiate up by fixing the numerator. Multiply (2 / 5y) * (3 / 3), to get (6 / 15y). you on the on the spot have: (((6 / 15y) - (3 / 15y)) / 2) / ((4 / 7) + (7 / 15y)) ((3 / 15y) / 2) / ((4 / 7) + (7 / 15y)) (3 / 30y) / ((4 / 7) + (7 / 15y)) (a million / 10y) / ((4 / 7) + (7 / 15y)) Now paintings on the denominator. Multiply (4 / 7) by (15y/15y) and (7 / 15y) by (7/7). (a million / 10y) / ((60y / 105y) + (40 9 / 105y)) (a million / 10y) / ((60y + 40 9) / 105y) Multiply the outer fraction by 10y / 10y. a million / ((10y (60y + 40 9)) / 105y) because we've a million interior the numerator, we are able to rewrite the outer fraction by using truth the reciprocal of the denominator. 105y / (10y (60y + 40 9)) 105y / (600y^2 + 490y) Divide by 5y/5y. 21 / (120y + ninety 8) finally, you're able to opt to ingredient out a 2 interior the denominator in case you want. 21 / (2(60y + 40 9)) 2. ((4 / 2z) + (5 / 6z)) / (7 / ((3 / z) - (4 / 6z))) Multiply (4 / 2z) by 3/3 and (3 / z) by 6/6. ((12 / 6z) + (5 / 6z)) / (7 / ((18 / 6z) - (4 / 6z))) (17 / 6z) / (7 / (14 / 6z)) Multiply the denominator by (6z / 14) / (6z / 14) (17 / 6z) / (7 (6z / 14)) (17 / 6z) / (42z / 14) (17 / 6z) / 3z 17 / (18z^2)

Answer 6

yes