Asked by Kristine
Help!! I am a home-school mom who did horrible in Honors Algebra2. My daughter has done her work and I am stumped on two questions.
Simplify. 6/x-6+3x/x+5-1/(x-6)(x+5)
A) 3x+5/x^2-x-30
B) 3x^2-12x+29/x^2-x-30
C) 3x^2+6x+5/x^2-x-30
D) 3x^2+5/x-6
Simplify the complex fraction.
(1/g+x+4/16)/(x+4/x^2+4/x)
A) x^4+4x^3/80x+64
B) x^4+4x^3/512x+2048
C) x^3+2x^2+4/80x+64
D) x^3+6x^2/80x+64
Thank you so much in advance
Simplify. 6/x-6+3x/x+5-1/(x-6)(x+5)
A) 3x+5/x^2-x-30
B) 3x^2-12x+29/x^2-x-30
C) 3x^2+6x+5/x^2-x-30
D) 3x^2+5/x-6
Simplify the complex fraction.
(1/g+x+4/16)/(x+4/x^2+4/x)
A) x^4+4x^3/80x+64
B) x^4+4x^3/512x+2048
C) x^3+2x^2+4/80x+64
D) x^3+6x^2/80x+64
Thank you so much in advance
Answers
Answered by
Kristine
Sorry its not a g in the second problem, it is an 8
Answered by
Reiny
You MUST include brackets to establish the correct order of operation.
I will assume you meant:
6/(x-6) + 3x/(x+5) - 1/( (x-6)(x+5) )
notice that the common denominator is (x-6)(x+5)
so ...
( 6(x+5) + 3x(x-6) - 1)/( (x-6)(x+5) )
= (6x + 30 + 3x^2 - 18x - 1)/((x-6)(x+5))
= (3x^2 - 12x + 29)/((x-6)(x+5))
which would be B, if it contained the brackets.
I will let you correct the 2nd question, since it clearly contains a typo.
I see it starts with (1/g ...
but I don't see any g in any of the answers.
I will assume you meant:
6/(x-6) + 3x/(x+5) - 1/( (x-6)(x+5) )
notice that the common denominator is (x-6)(x+5)
so ...
( 6(x+5) + 3x(x-6) - 1)/( (x-6)(x+5) )
= (6x + 30 + 3x^2 - 18x - 1)/((x-6)(x+5))
= (3x^2 - 12x + 29)/((x-6)(x+5))
which would be B, if it contained the brackets.
I will let you correct the 2nd question, since it clearly contains a typo.
I see it starts with (1/g ...
but I don't see any g in any of the answers.
Answered by
Kristine
Thanks for your help and yes I forgot the brackets sorry. It's supposed to be (1/8+(x+4/16))/(x+4/x^2)+(4/x)
Answered by
Reiny
ok, taking it the way you typed it
(1/8 + x + 4/16)/(x + 4/x^2 + 4/x)
= (1/8 + x + 1/4)/(x + 4/x^2 + 4/x)*(16x^2/(16x^2)
= (2x^2 + 32x^3 + 4x^2)/(16x^3 + 64 + 64x)
which does not come close to any of the given answers, so the original is missing some important brackets
(1/8 + x + 4/16)/(x + 4/x^2 + 4/x)
= (1/8 + x + 1/4)/(x + 4/x^2 + 4/x)*(16x^2/(16x^2)
= (2x^2 + 32x^3 + 4x^2)/(16x^3 + 64 + 64x)
which does not come close to any of the given answers, so the original is missing some important brackets
Answered by
Reiny
Was it perhaps
(1/8 + (x+4)/16 )/( (x+4)/x + 4/x ) ??
(1/8 + (x+4)/16 )/( (x+4)/x + 4/x ) ??
Answered by
Kristine
Yes sorry. I don't really know where to put most of the brackets
Answered by
Reiny
In that case
(1/8 + (x+4)/16 )/( (x+4)/x + 4/x )
= ( (2 + x+4)/16) / ( x+4 + 4)/x)
= (x+6)/16 * x/(x+8)
= x(x+6)/(x+8)
(1/8 + (x+4)/16 )/( (x+4)/x + 4/x )
= ( (2 + x+4)/16) / ( x+4 + 4)/x)
= (x+6)/16 * x/(x+8)
= x(x+6)/(x+8)
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