Question 1161302: Find three consecutive even integers such that the sum of 6 times the first, 5 times the second, and 4 times the third is 356.
Found 3 solutions by josgarithmetic, greenestamps, saw:
Answer by josgarithmetic(39620)   (Show Source):
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
Very often, a formal algebraic solution to a problem like this is made simpler by choosing the variable x as the middle number. In this problem, that leads to
The three numbers are 22, 24, and 26.
If a formal algebraic solution is not required, you can get to the solution quickly using a bit of estimation.
Since the numbers are three consecutive even integers, the sum of 6 times the first, 5 times the second, and 4 times the third is very close to 15 times the second.
Estimating 15 times an even number is about 356 leads immediately to 24 as the middle number, again making the three numbers 22, 24, and 26.
Answer by saw(34) (Show Source):
You can put this solution on YOUR website! Solution.
Let x = 1st number
Let x+2 = 2nd number
Let x+4 = 3rd number
Condition given
6x+5(x+2)+4(x+4) = 356
Distribute 5 through the parenthesis
6x+5x+10+4(x+4) = 356
Distribute 4 through the parenthesis
6x+5x+10+4x+16 = 356
Collect the like terms
15x+10+16 = 356
Add the numbers
15x+26 = 356
Move constant to the right side of the equation and change its sign
15x = 356-26
Subtract the numbers
15x = 330
Divide both sides of the equation by 15
x = 22
Use substitution
x+2 = 22+2 = 24
Use substitution
x+4 = 22+4 = 26
Three consecutive even integers are 22, 24, and 26.
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