2x / (y-2) + 2x / (2-y)
First get a common denominator: (y-2) (2-y)
Then you have:
2x(2-y) / (y-2)(2-y) + 2x(y-2) / (y-2)(2-y), summing and simplifying, you get:
(4x-4) / (y-2)(2-y)
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Assuming the equation is x2 - y2 - 6x - 2y = 60this is equivalent tox2 - 6x + 9 - y2 - 2y - 1 = 60 + 9 - 1= (x - 3)2 - (y + 1)2 = 68Assuming the equation is x2 - y2 - 6x - 2y = 60this is equivalent tox2 - 6x + 9 - y2 - 2y - 1 = 60 + 9 - 1= (x - 3)2 - (y + 1)2 = 68Assuming the equation is x2 - y2 - 6x - 2y = 60this is equivalent tox2 - 6x + 9 - y2 - 2y - 1 = 60 + 9 - 1= (x - 3)2 - (y + 1)2 = 68Assuming the equation is x2 - y2 - 6x - 2y = 60this is equivalent tox2 - 6x + 9 - y2 - 2y - 1 = 60 + 9 - 1= (x - 3)2 - (y + 1)2 = 68
5
Y1=2x^(2/3)+√(20-x²)-3 Y2=2x^(2/3)-√(20-x²)-3
Here it is: Y1=2x^(2/3)+√(20-x²)-3 Y2=2x^(2/3)-√(20-x²)-3
EQ1 x+y=8 EQ2 x2+y2=34 EQ1 x=8-y EQ1 and EQ2 combined gives (8-y)2+y2=34 simplify 64 -16y +y2+y2=34 simplify 2y2-16y+30=0 solve for -16y (2y-?)(y-!)=0 EQA EQ3 -2y!-y?=-16y simplify EQ3 2!+? =16 EQ4 ?!=30 EQ3 ?=16-2! Combined EQ3 and EQ4 gives (16-2!)!=30 simplify 8!-!2=15 Method of exhaustion 8!-!2 when !=0 -> 0 !=2 ->12 !=3->15 !=3 into EQ3 gives ?=16-2! gives ?=10 substitute into EQA gives (2y-10)(y-3)=0 solving gives y=5 or 3 solving for EQ1 x+y=8 then x=3 or 5 solving for EQ2 x2+y2=34 gives 9+25 for either combination so the larger number is 5