Question 114013: Three times the sum of two numbers exceeds twice their difference by five, while half the sum is two more than the difference. what are the two numbers?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Let the two numbers be x & y
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Write an equation for each statement:
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Three times the sum of two numbers exceeds twice their difference by five,
3(x+y) = 2(x-y) + 5
Arrange equation in the standard form
3x + 3y = 2x - 2y + 5
3x - 2x + 3y + 2y = 5
x + 5y = 5
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"while half the sum is two more than the difference."
.5(x+y) = x - y + 2
Get rid of the decimal, multiply by 2
x + y = 2x - 2y + 4
x - 2x + y + 2y = 4
-x + 3y = 4
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We can solve this by elimination:
x + 5y = 5
-x + 3y = 4
-------------adding eliminates x, find y
8y = 9
y = 9/8
y = 1.125
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Find x using x + 5y = 5:
x + 5(1.125) = 5
x + 5.625 = 5
x = 5 - 5.625
x = -.625
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Two non-integer solutions. See if they satisfy the original statement equations
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3(x+y) = 2(x-y) + 5
3(-.625 + 1.125) = 2(-.625 - 1.125) + 5
3(.5) = 2(-1.75) + 5
1.5 = -3.5 + 5
1.5 = 1.5 Confirms our solution
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You can check our solutions in:
.5(x+y) = x - y + 2
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