Math, ee-gads!!!!!!!!!!!!!!!?

Question

For what value(s) of the variable is each rational expression undefined?

10x/x-3
3/y(y^2-5y+6)

Simplfy the expression and state any restrictions on the variables.

16(x+1) / 30 (x+2)

the whole thing has me confused but it's the part about restrictions that really put me in a tail spin.

Thanks for any help you can give me!

Answer 1

basically, to find undefined, you want to make the denom. = 0.

in the first one, since the denominator is x-3, we can say that the expression is undefined when x - 3 = 0.
x - 3 = 0
add 3
x = 3
when x = 3, the expression is undefined.

for the second one, you have to find the roots and make them all = 0 as an OR statement. since you're multiplying them together, only one of them has to be zero for the expression to not work.

y(y^2-5y+6)

y(y+2)(y+3)

so we take each root by itself and make it = 0
dont forget the one outside the parentheses
y = 0

or

y+2=0
y= -2

or

y+3=0

y = -3

so y=0 or y= -2 or y= -3

the third one simplified, you distribute

16(x+1) / 30 (x+2)

16x + 16 / 30x + 60

8x + 8 / 15x + 30

since the bottom cannot equal zero, we will make it equal to zero to find the solution(s) that don't work.

15x + 30 = 0

subtract 30 from both sides

15x = -30

divide by 15

x = -2

which means x can NOT = -2 because then the denominator would be 0.
hope i helped!!

Answer 2

Hi,

A rational expression is undefined if its denominator is zero. To find when that is, set the denominator equal to zero and solve. That will give you the number(s) where the rational expression is undefined.

For 10x/x-3, let x - 3 = 0 and you get x = 3. That's where it is undefined.

For 3/y(y^2-5y+6), let y(y^2-5y+6) = 0 and factor it. It becomes y(y - 3)(y - 2) = 0 This solves to y = 0, 3, or 2. these are the values where this rational expression is undefined.

For the next expression, you would factor it if it could be factored. Then you would cancel. This becomes:

16(x+1)
------------ divide both numbers by 2 to get:
30 (x+2)

8(x+1)
------------ or
15(x+2)

8x +8
------------
15x+30

I hope this helps!! :-)

Answer 3

>> Math, ee-gads!!!!!!!!!!!!!!!?

Well of course it's "math". You're in the "mathematics" section. I sure as hell wasn't expecting you to post a problem here on how to speak Swahili.

Anyway, one rule we have in arithmetic is that you can't divide by zero. So these expressions are going to be undefined for values that make their denominator zero.

So in the first problem, which I'll assume is 10x/(x-3) and not (10x/x) - 3, the denominator will obviously be zero when x=3. So when x=3, it's undefined.

For 3 / [y(y^2-5y+6)], y=0 is one obvious answer. But if you factor the y^2 - 5y + 6 term, you'll find that there are two more.

Answer 4

They will be undefined when the denominator is 0.

10x/(x-3): undefined when x-3 = 0, i.e. x = 3.

3/[y(y^2-5y+6)]: undefined when
y(y^2-5y+6) = 0
<=> y(y-2)(y-3) = 0
<=> y = 0 or y-2 = 0 or y-3 = 0
<=> y = 0 or 2 or 3.

16(x+1) / 30(x+2)
= (16(x+2) - 16) / 30(x+2)
= 16/30(1 - 1/(x+2))
= 8/15 (1 - 1/(x+2))
The restrictions, again, are that a denominator cannot be 0, i.e. x+2 ≠ 0, so x ≠ -2.

Answer 5

The whole to this question is probably just to make you aware a fraction is undefined when the denominator is zero. So...

For the first one, x can't be 3.

For the second one, y can't be 2 or 3, since the denominator factors as (y - 2)(y - 3).

For the last one, x can't be -2.