Melissa giggled as she thought about 4 consecutive odd integers such that 5 times the sum of the first 2 was?

Question

10 less than 7 times the sum of the second and fourth. What were here four integers?

(Please show your work.)

Answer 1

4 consecutive odd integers are:

x, x+2, x+4, x+6

5(x+x+2) = -10 + 7(x+2+x+6)
10x+10 = -10 + 14x+56
-36 = 4x
-9=x

-9,-7,-5,-3 are the four numbers

Answer 2

If the first odd integer in the set is "n", then the following three odd integers must be n+2, n+4, and n+6. Five times the sum of the first two would be 5(n + n+2) = 5(2n+2) = 10(n+1). This is ten less than seven times the sum of the second and fourth, or 7(n+2 + n+6) - 10 = 7(2n+8)-10 = 14(n+4)-10. So:

10(n+1) = 14(n+4) - 10
10n+10 = 14n+56-10
10 = 4n+46
0 = 4n+36
0 = n+9

n = -9, so the integers are -9, -7, -5 and -3.

Answer 3

well, since they are consecutive odd integers, let's let the first one be x.

We know the other 3 will be x+2, x+4 and x+6

So...

5(x + x+2) = 7(x+2 + x+6) - 10
10x + 10 = 14x + 56 - 10
10x + 10 = 14x + 46
4x = -36
x = -9

so the 4 integers are: -9, -7, -5, and -3

Hope this helps!

Answer 4

4 consecutive odd integers could be:
x-4, x-2, x and x+2
hence:
5(2x-6)=7(2x)-10
10x-30=14x-10
-20=4x
x=-5

hence the numbers are -9,-7,-5,-3

Answer 5

But why did Melissa find the numbers -9, -7, -5, -3 so darn funny?

Answer 6

What he said ^^^^^^^^^^^^^^^^^^

The perosn below me did it wrong...its -36 not 36