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GRE Algebra If 2y^{−2} + 3y^{−1} − 14 = 0
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Updated on: 19 Nov 2015, 08:46
Updated on: 19 Nov 2015, 08:46
2
Expert Reply
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Question Stats:
83% (01:30) correct
16% (01:30) wrong based on 12 sessions
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If \(2*y^{-2} + 3*y^{-1} - 14 = 0\), which of the following could be the value of y?
A. 2
B. 1/2
C. 2/7
D. -1/2
E. -7
A. 2
B. 1/2
C. 2/7
D. -1/2
E. -7
Originally posted by RBornPowerScore on 19 Nov 2015, 07:15.
Last edited by RBornPowerScore on 19 Nov 2015, 08:46, edited 2 times in total.
Last edited by RBornPowerScore on 19 Nov 2015, 08:46, edited 2 times in total.
Re: GRE Algebra If y^-2 + 2y^-1 - 24 = 0
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19 Nov 2015, 07:17
19 Nov 2015, 07:17
1
Expert Reply
Show: :: Solution
Step 1: Apply the negative exponents rule.
\(a^{-n} = \frac{1}{a^{n}}\)
\(2(\frac{1}{y^2}) + 3(\frac{1}{y}) - 14 = 0\)
Step 2: Simplify.
\(2(\frac{1}{y^2}) + 3(\frac{1}{y}) = 14\)
\(\frac{1}{y}(2(\frac{1}{y}) + 3) = 14\)
\(\frac{2}{y} + 3 = 14y\)
Step 3: Plug the answer choices in the equation. Consider (A) . What happens when you plug in 2 ? The left-hand side, 2 + 3, is smaller and thus unequal to the right, 14 x 2. So move on to option (B) 1/2.
\(\frac{2}{(\frac{1}{2})} + 3 = 14(\frac{1}{2})\)
\(4 + 3 = 7\)
Thus, (B) is correct.
\(a^{-n} = \frac{1}{a^{n}}\)
\(2(\frac{1}{y^2}) + 3(\frac{1}{y}) - 14 = 0\)
Step 2: Simplify.
\(2(\frac{1}{y^2}) + 3(\frac{1}{y}) = 14\)
\(\frac{1}{y}(2(\frac{1}{y}) + 3) = 14\)
\(\frac{2}{y} + 3 = 14y\)
Step 3: Plug the answer choices in the equation. Consider (A) . What happens when you plug in 2 ? The left-hand side, 2 + 3, is smaller and thus unequal to the right, 14 x 2. So move on to option (B) 1/2.
\(\frac{2}{(\frac{1}{2})} + 3 = 14(\frac{1}{2})\)
\(4 + 3 = 7\)
Thus, (B) is correct.
gmatclubot
Re: GRE Algebra If y^-2 + 2y^-1 - 24 = 0 [#permalink]
19 Nov 2015, 07:17
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