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Question 531978: Find 3 consecutive numbers such that 4 times the first is equal to 7 more than the sum of the other two.
I do not understand how to work this problem. HELP!
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Find 3 consecutive numbers such that 4 times the first is equal to 7 more than the sum of the other two.
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3 numbers: x, (x+1), (x+2)
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Write an equation for the statement:
"numbers such that 4 times the first is equal to 7 more than the sum of the other two. "
4x = (x+1) + (x+2) + 7
4x = 2x + 3 + 7
4x = 2x + 10
4x - 2x = 10
2x = 10
x = 5
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The three numbers, 5, 6, 7
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Check that in the given statement:
"3 consecutive numbers such that 4 times the first is equal to 7 more than the sum of the other two. "
4(5) = 6 + 7 + 7
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