SOLUTION: Find four consecutive odd integers such that 10 more than 3 times the third is 40 less than the first?

Question 526891: Find four consecutive odd integers such that 10 more than 3 times the third is 40 less than the first?
Found 2 solutions by InfantileBear, MathTherapy:
Answer by InfantileBear(2) (Show Source): You can put this solution on YOUR website!
To start off, write the numbers such that they are all in terms of the same number, such as:
x,x+2,x+4,and x+6;since consecutive odd integers are two numbers apart.
Now, if you work backwards, you subtract 40 from the first number:
x-40
and it also says that three times the third plus 10 is the first minus 40, so;
3(x+4)+10=3x+22
so altogether
3x-22=x+40
if you solve, you will get x, which x=31
so the three consecutive odd integers are 31, 31+2, 31+4, and 31+6
or 31,33,35,37
Answer by MathTherapy(10858) (Show Source): You can put this solution on YOUR website!
Find four consecutive odd integers such that 10 more than 3 times the third is 40 less than the first?

Let the 1st integer be F

Then the others are: F + 2, F + 4, and F + 6

We then have: 3(F + 4) + 10 = F - 40

3F + 12 + 10 = F - 40

2F + 22 = - 40

2F = - 62

F, or 1st integer = , or - 31

The 4 consecutive odd integers are:

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