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a, b, and c are integers such that ab + c = 7, ac + b = 5, a
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18 Mar 2018, 06:17
18 Mar 2018, 06:17
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Question Stats:
76% (01:34) correct
23% (03:21) wrong based on 13 sessions
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a, b, and c are integers such that ab + c = 7, ac + b = 5, and a + b + c = 6. What is the value of abc ?
A. 2
B. 6
C. 9
D. 12
E. 18
A. 2
B. 6
C. 9
D. 12
E. 18
Drill 2
Question: 10
Page: 499
Question: 10
Page: 499
Re: a, b, and c are integers such that ab + c = 7, ac + b = 5, a
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18 Mar 2018, 07:31
18 Mar 2018, 07:31
1
Expert Reply
Explanation
By rearranging the first two equations, you obtain c = 7 – ab and b = 5 − ac.
By substituting these relationships into the third equation, a + (5 − ac) + (7 − ab) = 6 is obtained.
Manipulating the equation yields a − ac − ab + 12 = 6, or a − a (c + b) + 6 = 0. Since a + b + c = 6, b + c = 6 − a.
Substituting this into the previous equation yields \(a - a (6 - a) + 6 = a - 6a + a^2 + 6 = 0\), or \(a^2 - 5a + 6 = 0\).
Factoring and finding the roots gives you a = 2 or a = 3.
If a = 2—by substituting the value back into the equations—then b = 2 and c = 1. If a = 3, then b = 3 and c = 1. You are looking for abc, and either way, the value is 3 × 2 × 1 = 6. The answer is choice (B).
By rearranging the first two equations, you obtain c = 7 – ab and b = 5 − ac.
By substituting these relationships into the third equation, a + (5 − ac) + (7 − ab) = 6 is obtained.
Manipulating the equation yields a − ac − ab + 12 = 6, or a − a (c + b) + 6 = 0. Since a + b + c = 6, b + c = 6 − a.
Substituting this into the previous equation yields \(a - a (6 - a) + 6 = a - 6a + a^2 + 6 = 0\), or \(a^2 - 5a + 6 = 0\).
Factoring and finding the roots gives you a = 2 or a = 3.
If a = 2—by substituting the value back into the equations—then b = 2 and c = 1. If a = 3, then b = 3 and c = 1. You are looking for abc, and either way, the value is 3 × 2 × 1 = 6. The answer is choice (B).
Re: a, b, and c are integers such that ab + c = 7, ac + b = 5, a
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22 Mar 2018, 00:24
22 Mar 2018, 00:24
Solving this question is very time-consuming.
