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QOTD#21 If bc ≠ 0, and 3b + 2c = 18, then which of the
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02 Oct 2016, 09:12
02 Oct 2016, 09:12
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If bc ≠ 0, and 3b + 2c = 18, then which of the following is NOT a possible value of c?
A. \(5\frac{3}{5}\)
B. \(6\)
C. \(8\frac{2}{5}\)
D. \(9\)
E. \(12\)
A. \(5\frac{3}{5}\)
B. \(6\)
C. \(8\frac{2}{5}\)
D. \(9\)
E. \(12\)
Drill 1
Question: 5
Page: 283
Question: 5
Page: 283
Re: QOTD#21 If bc ≠ 0, and 3b + 2c = 18, then which of the
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02 Oct 2016, 09:18
02 Oct 2016, 09:18
Expert Reply
Explanation
This problem offers a good opportunity to plug in the answers—for simplicity’s sake, start with the integers. If c = 6, choice (B), then 3b + 2(6) = 18, so 3b + 12 = 18, 3b = 6, and b= 2. The only other requirement given is that bc ≠ 0, so 6 is, in fact, a possible value of c.
If c= 9, as in choice D, then 3b + 2(9) = 18, so 3b + 18 = 18, 3b = 0, and b = 0. A value of 0 for b would violate the given requirement, so 9 is NOT a possible value of c.
This problem offers a good opportunity to plug in the answers—for simplicity’s sake, start with the integers. If c = 6, choice (B), then 3b + 2(6) = 18, so 3b + 12 = 18, 3b = 6, and b= 2. The only other requirement given is that bc ≠ 0, so 6 is, in fact, a possible value of c.
If c= 9, as in choice D, then 3b + 2(9) = 18, so 3b + 18 = 18, 3b = 0, and b = 0. A value of 0 for b would violate the given requirement, so 9 is NOT a possible value of c.
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QOTD#21 If bc ≠ 0, and 3b + 2c = 18, then which of the
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13 Nov 2016, 08:12
13 Nov 2016, 08:12
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sandy wrote:
If bc ≠ 0, and 3b + 2c = 18, then which of the following is NOT a possible value of c?
A. \(5\frac{3}{5}\)
B. \(6\)
C. \(8\frac{2}{5}\)
D. \(9\)
E. \(12\)
A. \(5\frac{3}{5}\)
B. \(6\)
C. \(8\frac{2}{5}\)
D. \(9\)
E. \(12\)
Given: 3b + 2c = 18 [let's solve it for c]
Subtract 3b from both sides: 2c = 18 - 3b
Divide both sides by 2 to get: c = (18 - 3b)/2
Simplify: c = 18/2 - 3b/2
Simplify: c = 9 - 3b/2
NOTE: we're told that bc ≠ 0, which means b ≠ 0. If b ≠ 0, then 3b/2 CANNOT equal 0
So, it's impossible for c to equal 9, since that would mean that 3b/2 = 0, and we know that that's impossible.
Answer: D
