Asked by Crystal
Find three consecutive integers such that the sum of the first and second is three mose than three times the third.
When I did the problem,I got a negative answer. Does that seem correct? Did I make any mistakes?
When I did the problem,I got a negative answer. Does that seem correct? Did I make any mistakes?
Answers
Answered by
tchrwill
Let a = the first number, b = the second number and c = the third number.
Then, b = (a + 1) and c = (a + 2)
Therefore, a + (a + 1) = 3(a + 2) + 3 yielding 2a + 1 = 3a + 6 + 3 making a = -8 and the three numbers -8, -7 and -6.
-8 + -7 = 3(-6) + 3
-15 = -18 + 3 = -15.
Then, b = (a + 1) and c = (a + 2)
Therefore, a + (a + 1) = 3(a + 2) + 3 yielding 2a + 1 = 3a + 6 + 3 making a = -8 and the three numbers -8, -7 and -6.
-8 + -7 = 3(-6) + 3
-15 = -18 + 3 = -15.
Answered by
crystal
How do you get -8 as a?
Answered by
tchrwill
Let a = the first number, b = the second number and c = the third number.
Then, b = (a + 1) and c = (a + 2)
Therefore, a + (a + 1) = 3(a + 2) + 3 yielding 2a + 1 = 3a + 6 + 3.
.........2a + 1 = 3a + 9
.........2a + 1 - 9 = 3a
.........2a - 8 = 3a
............-8 = 3a - 2a
............-8 = a
-8 + -7 = 3(-6) + 3
-15 = -18 + 3 = -15.
Then, b = (a + 1) and c = (a + 2)
Therefore, a + (a + 1) = 3(a + 2) + 3 yielding 2a + 1 = 3a + 6 + 3.
.........2a + 1 = 3a + 9
.........2a + 1 - 9 = 3a
.........2a - 8 = 3a
............-8 = 3a - 2a
............-8 = a
-8 + -7 = 3(-6) + 3
-15 = -18 + 3 = -15.
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