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Question 765638: The sum of the squares of two numbers is 16 and twice of the first number take away three times of the second number is 3. Find the two numbers.
I think the two equations are:
Let the two numbers be x and y
x(square) + y(square) = 16
2x - 3y = 3
The answer is a decimal and i am not sure about the equations... if they are right or not... please help
Thanks
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The sum of the squares of two numbers is 16 and twice of the first number take away three times of the second number is 3. Find the two numbers.
I think the two equations are:
Let the two numbers be x and y
x(square) + y(square) = 16
2x - 3y = 3
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Solve the linear equation for "y":
y = (2x-3)/3
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Substitute into the top equation and solve fr "x":
x^2 + [(2x-3)/3]^2 = 16
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Get rid of the fraction:
9x^2 + (2x-3)^2 = 9*16
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9x^2 + 4x^2 - 12x + 9 = 9*16
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13x^2 - 12x - 15*9 = 0
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x = 3.7169 or x = -2.7939
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Solve for "y":
y = (2x-3)/3
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If x = 3.7169, y = 1.4779
If x = -2.7939, y = -2.8626
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Cheers,
Stan H.
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