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
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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.
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.
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