Asked by -Untamed-
Simlifying Rational Expressions
Reduce to lowest terms, stating the restrictions on the variable. I really don't understand the whole cancelling out the factor thing, and not cancelling out the terms. As I was having lots of trouble with this stuff this is the way I was told do solve an equation.
5a^3-15a^2/30a
5a^2(a-3)/30a
(5)(a)(a)(a-3)/(3)(2)(5)(a)
a(a-3)/6
a^2-3a/6
a cannot equal 0.
^ But even following it that way I still sort of get confused.
These are the ones I really need help on. And could you please just teach me the rules and stuff for Simplifying rational expressions and stating the restrictions. If you could show me a bunch of examples that would be great, thanks.
Questions - -"Stuck" on.
*a^2/a^2+a
*2-3/r-2
*Tried solving this one...
2b^2-18b/b(b-9)^2
2b(b-9)/b(b-9)(b-9)
I cancelled out the (b-9)'s and I cancelled out the b's( the one beside 2 and the one in the denominator, I wasn't sure whether they are considered as terms and if they should not be cancelled out, but it never made sense to not cancel them out so I did)
And I was left over with : 2/b-9
And the restriction part:
*Imagine the equal sign as a nonequal one
b-9 = 0
Added 9 to both sides
b = 9
Meaning b cannot equal nine. Is that the only restriction? Cause that's the only one I solved.
* t^2+4t+4/2t^2+10t+12
t^2+2t+2t+4/2t^2+6t+4t+12
t(t+2)2(t+2)/2t(t+3)4(t+3)
^ And I'm stuck :\
Reduce to lowest terms, stating the restrictions on the variable. I really don't understand the whole cancelling out the factor thing, and not cancelling out the terms. As I was having lots of trouble with this stuff this is the way I was told do solve an equation.
5a^3-15a^2/30a
5a^2(a-3)/30a
(5)(a)(a)(a-3)/(3)(2)(5)(a)
a(a-3)/6
a^2-3a/6
a cannot equal 0.
^ But even following it that way I still sort of get confused.
These are the ones I really need help on. And could you please just teach me the rules and stuff for Simplifying rational expressions and stating the restrictions. If you could show me a bunch of examples that would be great, thanks.
Questions - -"Stuck" on.
*a^2/a^2+a
*2-3/r-2
*Tried solving this one...
2b^2-18b/b(b-9)^2
2b(b-9)/b(b-9)(b-9)
I cancelled out the (b-9)'s and I cancelled out the b's( the one beside 2 and the one in the denominator, I wasn't sure whether they are considered as terms and if they should not be cancelled out, but it never made sense to not cancel them out so I did)
And I was left over with : 2/b-9
And the restriction part:
*Imagine the equal sign as a nonequal one
b-9 = 0
Added 9 to both sides
b = 9
Meaning b cannot equal nine. Is that the only restriction? Cause that's the only one I solved.
* t^2+4t+4/2t^2+10t+12
t^2+2t+2t+4/2t^2+6t+4t+12
t(t+2)2(t+2)/2t(t+3)4(t+3)
^ And I'm stuck :\
Answers
Answered by
Reiny
Questions - -"Stuck" on.
*a^2/a^2+a
= a^2/(a(a+1))
= a/(a+1) , a≠ 0
2-3/r-2 ????? , did you mean (2-r)/(r-2) ?
if so, then ...
= -1(r-2)/(r-2)
= -1 , r≠2
2b^2-18b/b(b-9)^2
I will assume you meant: (2b^2-18b)/( b(b-9)^2 )
= 2b(b-9)/((b-9)(b-9) )
= 2b/(b-9) , b ≠ 9
t^2+4t+4/2t^2+10t+12 , I will again assume you meant (t^2+4t+4)/(2t^2+10t+12)
= (t+2)(t+2)/(2 (t+2)(t+3) )
= (t+2)/(2t+6) , t ≠-3
In the last, it looks like you were factoring by decomposition.
I will do the bottom:
2t^2 + 10t + 12
= 2t^2 + 6t + 4t + 12
= 2t(t+3) + 4(2t+3) , you had these as being multiplied rather than added
= (t+3)(2t+4)
= (t+3)(2)(t+2)
= 2(t+3)(t+2)
*a^2/a^2+a
= a^2/(a(a+1))
= a/(a+1) , a≠ 0
2-3/r-2 ????? , did you mean (2-r)/(r-2) ?
if so, then ...
= -1(r-2)/(r-2)
= -1 , r≠2
2b^2-18b/b(b-9)^2
I will assume you meant: (2b^2-18b)/( b(b-9)^2 )
= 2b(b-9)/((b-9)(b-9) )
= 2b/(b-9) , b ≠ 9
t^2+4t+4/2t^2+10t+12 , I will again assume you meant (t^2+4t+4)/(2t^2+10t+12)
= (t+2)(t+2)/(2 (t+2)(t+3) )
= (t+2)/(2t+6) , t ≠-3
In the last, it looks like you were factoring by decomposition.
I will do the bottom:
2t^2 + 10t + 12
= 2t^2 + 6t + 4t + 12
= 2t(t+3) + 4(2t+3) , you had these as being multiplied rather than added
= (t+3)(2t+4)
= (t+3)(2)(t+2)
= 2(t+3)(t+2)
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