|
Question 188188: I am thinking of three consecutive negative numbers. If I multiply the first with the second and then subtract three times the third, the result is 57. What are the numbers? Show the algebra equation, and solve equation.
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
To start with, there is a problem with the way you stated the question. "...three consecutive negative numbers." doesn't make much sense unless those numbers are integers. Properly stated, the question should read, "I am thinking of three consecutive negative integers. Assuming that is the real case, the solution follows:
Let the first integer be x. Then the next consecutive integer is x + 1, and the next one is x + 1 + 1 = x + 2.
The first times the second is:
 = x^2 + x)
Three times the third is:
 = 3x + 6)
Subtract three times the third from the first times the second:
 = x^2 + x - 3x - 6 = x^2 - 2x -6)
And this is equal to 57:
Add -57 to both sides:
Factor:
(x - 9) = 0)
So:
or
Exclude the positive root because we are looking for three consecutive negative integers. Hence, the first negative integer is -7, the next is -6, and the third is -5:
 - 3(-5) = 42 + 15 = 57)
Answer checks.
Note that if the problem had not constrained the solution to negative integers, then 9, 10, and 11 would have worked: 90 - 33 = 57
John
|
|
|
| |
