Question 94849: Find three consecutive even integers whose sum is ten less than four times the smallest integer. Found 2 solutions by hard, ptaylor:Answer by hard(12) (Show Source):
You can put this solution on YOUR website! Since the answer numbers are consecutive, they will be equal to -
x, x+2, and x+4
Your entire equation should be x + x + 2 + x + 4 = 4x - 10
[Combine like terms]
3x + 6 = 4x - 10
[Subtract 3x from both sides]
6 = x - 10
[Add 10 to both sides]
x = 16 [the smallest integer in the set]
x + 2 = 18
x + 4 = 20
So, your three consecutive even integers should be 16, 18, and 20.
But, always check your answer to make sure it's correct.
16 + 18 + 20 = 4(16) - 10
54 = 64 - 10
54 = 54
So, your answer is correct!
You can put this solution on YOUR website!
Let x=smallest integer
Then x+2=next integer
And x+4=third integer
Now we are told that:
x+x+2+x+4=4x-10 start collecting like terms
3x+6=4x-10 subtract 4x and also 6 from both sides
3x-4x+6-6=4x-4x-10-6 finish collecting like terms
-x=-16 divide both sides by -1
x=16 smallest
x+2=16+2=18 2nd integer
x+4=16+4=20----third integer
CK
16+18+20=4*16-10
54=64-10
54=54
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