You can put this solution on YOUR website! Please help me sovle this system.
y = -1/(2x) +3 ----(1) and y = (6x-5) ----(2)
Putting (2) in (1) that is substituting for y in (1) we have
y = -1/(2x) +3 ----(1) giving
(6x-5) = -1/(2x) +3
Multiplying through out by (2x)
2x(6x-5) = - 1 +3X(2x)
12x^2-10x = -1+6x
12x^-10x-6x+1=0 (transferring 6x and -1 from the right to the left)
(change side then change sign)
12x^2 - 16x +1 = 0
Solving using formula x = [-b +orminus(sqrt(b^2-4ac)]/2a
where here a = 12, b = -16 and c = 1
x = [-(-16) + or minus(sqrt(256-48)]/(2X12)
x= [16 + or minus (sqrt 208)]/24
sqrt(208) = sqrt(16X13) = 4sqrt(13)
Therefore x = {16+ 0r minus [4sqrt(13)]}/(4X6)
cancelling 4
x = {4+ 0r minus [sqrt(13)]}/6
That is x =(1/6)[4+ sqrt(13)] and x= (1/6)[4- sqrt(13)]
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