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Algebra 13 Online
OpenStudy (anonymous):

xy=240, find x-y when x+y=32 and x>y can some one provide the solution

OpenStudy (anonymous):

The two conditions imply\[\frac{ 240 }{ x } = 32 - x\] \[x \in [ 12, 20 ] => (x,y) \in [ (12, 20), (20,12) ]\]

OpenStudy (anonymous):

thanks for taking time to respond, can you please explain as I did not quite understand

OpenStudy (anonymous):

Solve xy = 240 for y to get \[y = \frac{ 240 }{ x }\] Solve x + y = 32 for y to get \[y = 32 - x\] Put these together to get the expression that contains only 'x', as in my first response. Simplify it to get \[240 = x ( 32 - x ) = 32x - x ^{2}\] Rearrange this to get a quadratic equation in standard form.\[x ^{2} - 32x + 240 = 0\] Solve this to obtain two values for 'x'. For each of these values of 'x' use one of the expressions for'y' to obtain the corresponding value for 'y'. Now choose the (x,y) pair where x>y.

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