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29 May 2015, 20:10
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If m does not equal 2 or -3, then \(\frac{(m - 2) - \frac{3(m - 2)}{m} }{m^2 + m - 6}=\)?
A. \( \frac{m -3}{m^2 + 3m} \)
B. \( \frac{1}{m} \)
C. \( \frac{m - 2}{m^2 - 2m} \)
D. \( \frac{(m - 6)(m + 1)}{(m + 3)(m - 2)} \)
E. \( \frac{-3m + 6}{m^2 + 3m} \)
Manhattan m-2.jpg [ 22.45 KiB | Viewed 7612 times ]
A. \( \frac{m -3}{m^2 + 3m} \)
B. \( \frac{1}{m} \)
C. \( \frac{m - 2}{m^2 - 2m} \)
D. \( \frac{(m - 6)(m + 1)}{(m + 3)(m - 2)} \)
E. \( \frac{-3m + 6}{m^2 + 3m} \)
Attachment:
Manhattan m-2.jpg [ 22.45 KiB | Viewed 7612 times ]
29 May 2015, 20:36
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There'd be a few ways to carry out the algebra, for example:
We can factor the denominator: m^2 + m - 6 = (m -2)(m + 3)
We can also factor out (m-2) from each of the two terms in the numerator.
Then we'll get (m-2) on top and bottom of the fraction, so we can cancel. If we then multiply the numerator and denominator by m in order to get rid of the fraction within the numerator, we'll get something that looks like one of the answer choices:
\(\begin{align}\\
\dfrac{ (m - 2) - \frac{3(m-2)}{m}}{m^2 + m - 6} &= \dfrac{ (m - 2) \left(1 - \frac{3}{m} \right) }{ (m -2)(m + 3)} \\\\
&= \dfrac{1 - \frac{3}{m}}{m + 3} \\\\
&= \left( \dfrac{1 - \frac{3}{m}}{m + 3} \right) \left( \frac{m}{m} \right) \\ \\
&= \frac{m - 3}{m^2 + 3m}\\
\end{align}\)
We can factor the denominator: m^2 + m - 6 = (m -2)(m + 3)
We can also factor out (m-2) from each of the two terms in the numerator.
Then we'll get (m-2) on top and bottom of the fraction, so we can cancel. If we then multiply the numerator and denominator by m in order to get rid of the fraction within the numerator, we'll get something that looks like one of the answer choices:
\(\begin{align}\\
\dfrac{ (m - 2) - \frac{3(m-2)}{m}}{m^2 + m - 6} &= \dfrac{ (m - 2) \left(1 - \frac{3}{m} \right) }{ (m -2)(m + 3)} \\\\
&= \dfrac{1 - \frac{3}{m}}{m + 3} \\\\
&= \left( \dfrac{1 - \frac{3}{m}}{m + 3} \right) \left( \frac{m}{m} \right) \\ \\
&= \frac{m - 3}{m^2 + 3m}\\
\end{align}\)
29 May 2015, 21:06
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Another strategy that will work here is to pick a number for m, and work out what the expression equals for that value of m. So for example, you could plug in m = 1. You'd then find, after some annoying calculation, that the expression equals -1/2.
Then when you plug m=1 into each answer choice, the right answer will also need to be equal to -1/2. If you do that, only answer A gives you -1/2, so it must be correct. One potential danger with this approach is that two answers, by pure coincidence, might both give you -1/2 for your chosen value of m. Then you need to pick another value of m and redo all the work to choose between those two answers.
I would personally find all that number-plugging a lot slower than doing the algebra, but for some test takers it will be faster; it really depends how comfortable you are with the algebraic manipulation the question requires. It's at least worth knowing as a fallback strategy if you don't see any algebraic simplification quickly. To use that strategy well, you need to be able to zero in quickly on a number that will be convenient to calculate with. Often 0 and 1 are good choices, when they're permitted (in the question above, even though the question fails to mention the fact, m cannot be 0, because 'm' appears in a denominator). But in some cases, 0 and 1 are bad choices. So in this question:
\(\dfrac{(m - 72)^2 + (m - 72)}{(m - 71)} =\)
\(\begin{align}\\
&\text{A) } m^2 - 72^2 \\\\
&\text{B) } m^2 - 144m - 72 \\\\
&\text{C) } m - 71 \\\\
&\text{D) } m - 72 \\\\
&\text{E) } m - 73 \\
\end{align}\)
you'd be making your life more difficult than necessary by plugging in m=1, because then you'd get things like (-71)^2 in the numerator. It would be better to plug in something like m=73, so that the expressions (m-72) and (m-71) in the question turn out to be equal to easy numbers. Notice though that some answer choices (B especially) are going to be a bit irritating to plug numbers into no matter what number you choose.
I'd still use algebra in the question above - if you notice you can factor out (m-72) from both of the terms in the numerator you get the answer almost instantly:
\(\begin{align}\\
\frac{(m - 72)^2 + (m - 72)}{(m - 71)} &= \frac{(m-72)[ (m - 72) + 1]}{m-71} \\\\
&= \frac{(m-72)(m-71)}{m - 71} \\\\
&= m - 72\\
\end{align}\)
Then when you plug m=1 into each answer choice, the right answer will also need to be equal to -1/2. If you do that, only answer A gives you -1/2, so it must be correct. One potential danger with this approach is that two answers, by pure coincidence, might both give you -1/2 for your chosen value of m. Then you need to pick another value of m and redo all the work to choose between those two answers.
I would personally find all that number-plugging a lot slower than doing the algebra, but for some test takers it will be faster; it really depends how comfortable you are with the algebraic manipulation the question requires. It's at least worth knowing as a fallback strategy if you don't see any algebraic simplification quickly. To use that strategy well, you need to be able to zero in quickly on a number that will be convenient to calculate with. Often 0 and 1 are good choices, when they're permitted (in the question above, even though the question fails to mention the fact, m cannot be 0, because 'm' appears in a denominator). But in some cases, 0 and 1 are bad choices. So in this question:
\(\dfrac{(m - 72)^2 + (m - 72)}{(m - 71)} =\)
\(\begin{align}\\
&\text{A) } m^2 - 72^2 \\\\
&\text{B) } m^2 - 144m - 72 \\\\
&\text{C) } m - 71 \\\\
&\text{D) } m - 72 \\\\
&\text{E) } m - 73 \\
\end{align}\)
you'd be making your life more difficult than necessary by plugging in m=1, because then you'd get things like (-71)^2 in the numerator. It would be better to plug in something like m=73, so that the expressions (m-72) and (m-71) in the question turn out to be equal to easy numbers. Notice though that some answer choices (B especially) are going to be a bit irritating to plug numbers into no matter what number you choose.
I'd still use algebra in the question above - if you notice you can factor out (m-72) from both of the terms in the numerator you get the answer almost instantly:
\(\begin{align}\\
\frac{(m - 72)^2 + (m - 72)}{(m - 71)} &= \frac{(m-72)[ (m - 72) + 1]}{m-71} \\\\
&= \frac{(m-72)(m-71)}{m - 71} \\\\
&= m - 72\\
\end{align}\)
