6y^2 - 8xy = 9y - 3x - 2x^2

Question

I need to use the completing the square method and follow this instruction: solve for y in terms of x

Answer 1

6y²-8xy-9y+3x+2x²=0,

6y²-y(8x+9)+2x²+3x=0.

Consider the quadratic ay²-by+c=0,

y²-by/a=-c/a,

y²-by/a+b²/(4a²)=b²/(4a²​)-c/a,

(y-b/(2a))²=b²/(4a²​)-c/a,

y-b/(2a)=±√(b²/(4a²​)-c/a).

When a=6, b=8x+9, c=2x²+3x:

y-(8x+9)/12=±√((64x²+144x+81)/144-(2x²+3x)/6),

y-(8x+9)/12=±(1/12)√(64x²+144x+81-48x²-72x),

y-(8x+9)/12=±(1/12)√(16x²+72x+81)=±(1/12)(4x+9),

y=(8x+9+4x+9)/12=(12x+18)/12=x+3/2 or (8x+9-4x-9)/12=x/3.

CHECK: 6(y-x-3/2)(y-x/3)=(2y-2x-3)(3y-x)=6y²-8xy+2x²-9y+3x=0,

6y²-8xy=9y-3x-2x².