The sum of three consecutive even integers (3 even integers in a row) is 120.
Use the following setup and given equation to find your solution:
1st integer = x
2nd integer = x+2
3rd integer = x+4
Equation: x+(x+2)+(x+4)=120
Solve the equation above. What is the largest of the three integers?
A. 38
B. 42
C. 44
D. 38,42,44
To solve the equation x + (x + 2) + (x + 4) = 120, we can combine like terms:
3x + 6 = 120
Next, we can subtract 6 from both sides of the equation:
3x = 114
Finally, we divide both sides of the equation by 3 to solve for x:
x = 38
Therefore, the first integer is 38, the second integer is 38 + 2 = 40, and the third integer is 38 + 4 = 42.
The largest of the three integers is 42.
Therefore, the answer is B. 42.