Four times the greatest of three consecutive odd
integers exceeds three times the least by 31.
What is the greatest of the three consecutive
odd integers?

Respuesta :

Answer:

The largest is 25.

Step-by-step explanation:

If the first integer is x, the second is x+1 and the third is x+2.

When you multiply the largest by 4: (x+2)*4

It is 31 more than 3 times the smallest: (x+2)*4 = (x)*3+31

When you simplify you get : 4x+8 = 3x+31

Which means: x = 23

But since you want the largest you get: 23+2 = 25

The greatest of the three consecutive odd integers will be  [tex]19[/tex].

What are integers ?

Integers are whole numbers that can be positive, negative and zero.

We have,

Let first integer [tex]= x[/tex]

Second integer [tex]= x+2[/tex]

Third integer  [tex]= x+4[/tex]

Now,

According to the question,

[tex]4(x+4) = 3(x)+31[/tex]

Simplifying;

[tex]4x+16=3x+31[/tex]

Taking [tex]x[/tex] variable on one side and constant on the other;

[tex]4x-3x=31-16[/tex]

[tex]x=15[/tex]

So,

Now substitute the value of [tex]x[/tex]  in every integer's equation,

i.e.

First integer [tex]= 15[/tex]

Second integer [tex]= 17[/tex]

Third integer  [tex]= 19[/tex]

So, the greatest of the three consecutive odd integers is [tex]19[/tex].

Hence, we can say that the greatest of the three consecutive odd integers will be  [tex]19[/tex].

To know more about integer click here

https://brainly.com/question/15276410

#SPJ2