Question

7x+3/x2-8x+15+3x/x-5=1/3-x

Answer 1

Combine like terms
(3/x^2) + 7x - 8x + 15 + 3x/x - 5 = (1/3) - x
The x will cancel out of 3x/x term leaving just x. Move all x's to left of equal sign and all constants to the right.
(3/x^2)+7x-8x+x= -15-3+5+(1/3)
(3/x^2)=-13+(1/3)
Convert -13 to a fraction - -39/3 and add to 1/3
(3/x^2) = -38/3
Multiply both sides by x^2.
3 = (-38/3)x^2
Multiply both sides by (-3/38)
(-9/38) =x^2
Take square root of both sides
Sq rt (-9/38) = x
Bring an i out from under the sq rt to get rid of the negative sign.
3i/(sq rt 38) = x
Multiply top and bottom by sq rt 38 to get it off the denominator.
3i(sq rt 38)/38 = x