System of two equations problem
- Thread starter bbhoney
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- Apr 12, 2005
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That is a little painful.
Rule #1 - Never do it the hard way unless someone is forcing you to do so.
4x^2 + 4y^2 = 82 ==> 2x^2 + 2y^2 = 41 That's got to be easier.
(5-x)(5-x) = (5-x)^2 Memorize this (5-x)^2 = 25 - 10x + x^2 Don't EVER do it in four pieces again. Square the first, 2*first*last, square the last.
Question #1 - How are you in "Intermediate" algebra and the Quadratic formula does not spring immediately to mind? Did you try factoring? How many quadratic equations have you solved in your lifetime? More than just a few, I suspect.
8x^2-40x+18=0
First, refer to "Rule #1".
4x^2 - 20x + 9 = 0
This factors:
(2x 9)(2x 1) = 0
I'll let you supply the correct signs.
P.S. Just an interesting note for you to explore. It doesn't necessarily help too much in this particular problem, but you may wish to keep it in mind for future reference.
If x+y = 5, then (x+y)^2 = x^2 + 2xy + y^2 = 5^2 = 25 OR x^2 + y^2 = 25 - 2xy OR 2x^2 + 2y^2 = 50 - 4xy
This could be very useful, since we know already that 2x^2 + 2y^2 = 41, giving 41 = 50 - 4xy OR 4xy = 9
With x + y = 5 and 4xy = 9 OR y = 9/(4x), your substitution now leads quite directly to the same quadratic equation.
Really, just something to think about. Don't lose too much sleep over it.
Rule #1 - Never do it the hard way unless someone is forcing you to do so.
4x^2 + 4y^2 = 82 ==> 2x^2 + 2y^2 = 41 That's got to be easier.
(5-x)(5-x) = (5-x)^2 Memorize this (5-x)^2 = 25 - 10x + x^2 Don't EVER do it in four pieces again. Square the first, 2*first*last, square the last.
Question #1 - How are you in "Intermediate" algebra and the Quadratic formula does not spring immediately to mind? Did you try factoring? How many quadratic equations have you solved in your lifetime? More than just a few, I suspect.
8x^2-40x+18=0
First, refer to "Rule #1".
4x^2 - 20x + 9 = 0
This factors:
(2x 9)(2x 1) = 0
I'll let you supply the correct signs.
P.S. Just an interesting note for you to explore. It doesn't necessarily help too much in this particular problem, but you may wish to keep it in mind for future reference.
If x+y = 5, then (x+y)^2 = x^2 + 2xy + y^2 = 5^2 = 25 OR x^2 + y^2 = 25 - 2xy OR 2x^2 + 2y^2 = 50 - 4xy
This could be very useful, since we know already that 2x^2 + 2y^2 = 41, giving 41 = 50 - 4xy OR 4xy = 9
With x + y = 5 and 4xy = 9 OR y = 9/(4x), your substitution now leads quite directly to the same quadratic equation.
Really, just something to think about. Don't lose too much sleep over it.