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Question 515466: The sum of 3 numbers is 29. Twice the first number is 18 more than three times the second number. The difference of the third and second numbers is one more than the first numbers. Find the numbers.
Answer by oberobic(2304) (Show Source):
You can put this solution on YOUR website! x + y + z = 29
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Twice the first number is 18 more than 3 times the second number.
2x = 3y +18
so
2x - 3y = 18
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The third number minus the second number is 1 more than the first number.
z - y = x+1
so
-x -y +z = 1
.
x + y + z = 29
2x -3y = 18
-x -y +z = 1
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Add the first and third equation
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x + y + z = 29
-x -y +z = 1
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2z = 30
z = 15
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substitute z=15
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x +y + z = 29
x + y = 14
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Recall
2x = 3y + 18
2x -3y = 18
and
x + y = 14
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Multiply x+y=14 by 2 and subtract from 2x-3y=18
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2x -3y = 18
2x +2y = 28
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-5y = -10
y = 2
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x+y = 14
x = 12
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Answer: The numbers are 2, 12, and 15.
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Always check your answer!
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Does 2x = 3y + 18?
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2*12 = 24
3*2 = 6
6+18 = 24
Correct.
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Is the third number minus the second number is 1 more than the first number?
z = 15
y = 2
15-2 = 13
x = 12
Yes.
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Done.
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