Find four consecutive odd integers such that the sum of the first three exceeds the fourth by 18?

Let, numbers = x,(x+2),(x+4),(x+6)

A.T.Q,

x+x+2+x+4 = x+6+18

3x+6 = x+24

3x-x= 24-6

2x = 18

x = 9

Numbers would be 9, 11, 13, 15

9,11,13,15

Step-by-step explanation:

Let our 4 numbers be

-n

-n+2

-n+4

-n+6

first, we have to write an expression and we gradually simplify.

n+n+n+2+4=n+6+18

combine like terms:

3n+6==n+24

subtract from both parts of the equation.

3n-n=24-6

now are finally finishing our problem.

2n=18

n=9

Now that we know our first integer, we need to consecute.

9,11,13,15

0.6, 3/5, and 6% are the 3 same portion.

but, when they're all changed to a decimal they would equal to 0.6.

hope this , good luck!

(0.2)+(-8)=0.3+-3+0.3m is the equation i guess? ?

step-by-step explanation: