# The sum of three integers is 66. The second is 2 more than the first, and the third is 4 more than twice the first. What are the integers?

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Let the integers be n, m and k

==> n + m + k = 66

given the second = 2 more than the first:

=> m = 2 + n........(1)

given the third i= 4 more that twice the first:

==> k = 4+ 2n ........(2)

Now substitute in the equation:

n + m + k = 6

==> n + (2+ n) + (4+ 2n) = 66

Now combine like terms:

==> 4n + 6 = 66

Now subtract 6 from both sides:

==> 4n = 60

Noe divide by 4:

==> n = 15

==> m= 15+ 2 = 17

==> k= 4 + 2*15 = 34

**Then the three integers are: 15, 17, and 34**

**To confirm:**

**15+ 17 + 34 = 66**

The smallest term is x.

x+ y +x = 66 (1)

y = x + 2 (2)

z = 2x + 4 (3)

Now, we'll substitute (2) and (3) in (1):

x + x + 2 + 2x + 4 = 66

We'll combine like terms:

4x + 6 = 66

We'll subtract 6 both sides:

4x = 66-6

4x = 60

We'll divide by 4:

x = 15

The second term, y is:

y = x + 2

y = 15 + 2

y = 17

The 3rd term is:

z = 2x+4

z = 2*15 + 4

z = 34

**The integer numbers are: 15 , 17 and 34.**

Let the first integer be x.

The 2nd integer is 2 more than the first. So the second integer = x+2.

The third integer is 4 more than twice the first. So the third integer = 4+2x.

Also the sum of the three numbers x , 2+x and 4 +2x is 66.

Therefore x+2+x+4+2x = 66. Or

4x+6 = 66.

4x = 66-6

4x= 60

x = 60/4 = 15.

Therefore the three numbers are:

x = 15

x+2 = 15+2 = 17.

2x+4 = 15*2+4 = 34.

Verification: sum of the 3 numbers = 15+17+34 = 66.

We have that the sum of the three integers is 66.

Let them be a, b and c.

Now the second is 2 more than the first

=> b = a + 2

And the third is 4 more than twice the first

=> c = 2a + 4

Therefore we can rewrite a + b + c = 66

=> a + 2 + 2a + 4 + a = 66

=> 4a + 6 = 66

=> 4a = 60

=> a = 15

Therefore b = 17

and c = 34

**The required numbers are 15 , 17 and 34.**