SOLUTION: The sum of two numbers is five times their difference. If one number exceeds the other by 7, find the numbers.

Question 521871: The sum of two numbers is five times their difference. If one number exceeds the other by 7, find the numbers.
Found 2 solutions by stanbon, ssm@123:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The sum of two numbers is five times their difference. If one number exceeds the other by 7, find the numbers.
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Equations:
x+y = 5(x-y)
x = y + 7
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Substitute for "x" and solve for "y":
y+7+y = 5(y+7-y)
2y + 7 = 35
2y = 28
y = 14 (one of the numbers)
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Solve for "x":
x = y + 7
x = 14+7
x = 21 (the other number)
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Cheers,
Stan H.
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Answer by ssm@123(2) (Show Source): You can put this solution on YOUR website!
let the first number be - x
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let the second number be - x+7
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the sum of the 2 numbers is - 5(x+(x+7))
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required equation -

x+(x+7) = 5*(x-(x+7)
x+x+7 = 5*(x-x+7)
2x+7 = 5x-5x+35
2x+5x-5x = 35-7
7x-5x = 28
2x = 28
x = 28/2
x = 14
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So, the first number = x = 14
the second number = x+7 = 14+7
= 21
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Therefore, the numbers are 7 and 21
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SOLVED BY -
SRAVYA SRI
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