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Mathematics 16 Online
OpenStudy (anonymous):

If x2 + y2 = 21 and 2x - 2y = 2, then what is x? If there are two possible answers, then whats the larger of the 2

OpenStudy (misty1212):

hi!!

OpenStudy (misty1212):

\[x^2+y^2=21\\ 2x-2y=2\] or easier \[x-y=1\] so \[y=x-1\]

OpenStudy (misty1212):

you can substitute \(x-1\) for \(y\) in the first line and solve \[x^2+(x-1)^2=22\] for \(x\)

OpenStudy (misty1212):

you ok from that point on?

OpenStudy (anonymous):

can you help me through it? @misty1212

OpenStudy (misty1212):

ok we have to go ahead and square the second term \[x^2+(x-1)^2=x^2=(x-1)(x-1)\\ =x^2+x^2-2x+1\\ 2x^2-2x+1\]

OpenStudy (anonymous):

okay I got that then what?

OpenStudy (misty1212):

\[2x^2-2x+1=22\\ 2x^2-2x-21=0\] and if you are lucky it factors

OpenStudy (anonymous):

I use the quadratic eqn right? @misty1212

OpenStudy (misty1212):

hmm hold on usually these are cooked up to give nice answers, this one is ugly let me see if i made a mistake

OpenStudy (misty1212):

ooh yesssss it was 21 not 22!

OpenStudy (anonymous):

okay thank you! @misty1212

OpenStudy (misty1212):

\[2x^2-2x+1=21\\ 2x^2-2x-20=0\]

OpenStudy (misty1212):

still not that nice though \[x^2-x-10=0\]

OpenStudy (anonymous):

then i do quadratic eqn? @misty1212

OpenStudy (misty1212):

yes there is no avoiding it

OpenStudy (anonymous):

what did you get for it? @misty1212

OpenStudy (misty1212):

\[x=\frac{1\pm\sqrt{41}}{2}\] i think

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