SOLUTION: Please help! Find four consecutive integers such that 5 times the opposite of the first is 5 greater than the product of -3 and the sum of the third and fourth. Thanks.

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Question 1165568: Please help!
Find four consecutive integers such that 5 times the opposite of the first is 5 greater than the product of -3 and the sum of the third and fourth.
Thanks.
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
system%28n%2C+n%2B1%2C+n%2B2%2C+n%2B3%29

5%28-n%29=5%2B%28-3%29%28%28n%2B2%29%2B%28n%2B3%29%29
--
-5n=5-3%28n%2B2%29-3%28n%2B3%29
-5n=5-3n-6-3n-9
-5n=5-6n-15
-5n=-6n-10
5n=6n%2B10
n=-10-----------and you can finish from this.
.

Answer by MathTherapy(10551) About Me  (Show Source):
You can put this solution on YOUR website!

Please help!
Find four consecutive integers such that 5 times the opposite of the first is 5 greater than the product of -3 and the sum of the third and fourth.
Thanks.
Let first number be F

Then 2nd, 3rd, and 4th are: F + 1, F + 2, and F + 3, respectively 

We then get: 5(- F) = - 3(F + 2 + F + 3) + 5

- 5F = - 3(2F + 5) + 5

- 5F = - 6F - 15 + 5

- 5F = - 6F - 10

- 5F + 6F = - 10

First number or highlight_green%28matrix%281%2C3%2C+F%2C+%22=%22%2C+-+10%29%29

You should now be able to find the other 3!