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. \(12\)
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/2NOTE: 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
_________________
Brent Hanneson - founder of Greenlight Test Prep