Greetings!
To start this problem, let's first assign a variable for the missing, consecutive odd numbers. Since they are consecutive and odd, we add two.
Proof: 3-1=2, 5-3=2
The first, consecutive, odd number: [tex] x [/tex]
The second, consecutive, odd number: [tex] x+2 [/tex]
The third, consecutive, odd number: [tex] x+4 [/tex]
The fourth, consecutive, odd number: [tex] x+6 [/tex]
The sum of the values are equal to 3 times the sum of the first two numbers, of which this is equal to 35 less than the fourth number. Let's create an equation to simplify this:
[tex] 3((x)+(x+2))=(x+6)-35 [/tex]
Complete the operations inside the parenthesis:
[tex] 3(2x+2)=(x+6)-35 [/tex]
Distribute the parenthesis (utilizing the distributive property)
[tex] (((2x)(3))+((2)(3)))=(x+6)-35 [/tex]
[tex] (6x+6)=(x+6)-35 [/tex]
Simplify both sides:
[tex]6x+6=x-29[/tex]
Add -6 and -x to both sides of the equation:
[tex](6x+6)+(-6)+(-x)=(x-29)+(-6)+(-x)[/tex]
[tex]5x=-35[/tex]
Divide both sides of the equation by 5:
[tex] \frac{5x}{5}=\frac{-35}{5} [/tex]
[tex] x=-7 [/tex]
If [tex]x[/tex] is equal to -7:
[tex] x+2=-5 [/tex]
[tex] x+4=-3 [/tex]
[tex] x+6=-1 [/tex]
The four numbers are:
[tex] \boxed{-7,-5,-3,-1} [/tex]
I hope this helps!
-Benjamin
