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No M's in Algebra

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by edvhou812 » Wed Jul 13, 2011 10:01 pm
I can't seem to find this question addressed in the forum. I'm trying to get stronger in algebra, and plugging in is too much work for this question. Different types of algebraic solution methods are appreciated. For some reason, I keep picking "E" as the incorrect answer. Note: I had to press "refresh" for the full jpeg to appear.
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by amit2k9 » Wed Jul 13, 2011 10:15 pm
for denominator reduction gives (m-2)(m+3)

for numerator = (m-2)(m-3)/m

thus ratio = m-3/m*(m+3) = A
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by tpr-becky » Thu Jul 14, 2011 9:53 am
I would still recommend plugging in on this problem however you could do it algebraically by first factoring the numerator and denominator to get
((m-2)(1-3/m))/((m-2)(m+3))
Then you can cancel the m - 2 to get (1- 3/m)/(m+3)
But none of the answers are in this form and this equation could be tricky so try factoring the answers instead.
Answer A (m-3)/(m^2+ 3m) can be factored to (m(1-3/m))/(m(m+3)) cancel out the m's and you have a match for the above. So the correct answer is A.
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by kitg » Thu Jul 14, 2011 8:10 pm
Best thing is to simplify these kind of expressions
numerator reduces to ( (m-2)*(m-3) ) /m

denominator reduces to (m+3)(m-2)

so numerator by denominator will equal to (after cancelling the common thing (m-2) )
= (m-3) / (m*(m+3 )) => A
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by MBA.Aspirant » Fri Jul 15, 2011 12:03 am
edvhou812 wrote:I can't seem to find this question addressed in the forum. I'm trying to get stronger in algebra, and plugging in is too much work for this question. Different types of algebraic solution methods are appreciated. For some reason, I keep picking "E" as the incorrect answer. Note: I had to press "refresh" for the full jpeg to appear.
Algebra:

(m-2) - 3(m-2)/m/ m^2+m-6

m(m-2) - 3(m-2)/m / m^2+m-6

(m-2) (m-3)/ m(m+3) (m-2)

m-3/ m(m+3)

m-3/ m^2+3m

Pluggin:

(m-2) - 3(m-2)/m/ m^2+m-6

say m =1

(1-2) - 3 (1-2)/1 / 1^2 +1 -6

2/-4 = -1/2

plugin 1 in A

1-3/1+3 = -2/4 = -1/2
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