I have equation

$\displaystyle y=\frac{2x-5}{3x+2}$

Please check my working.(I started expanding the right hand side of the numerator first , i hope you can follow.

$\displaystyle =\frac{[(3x+2)^2(2)]-[(2x-5)[(3x+2)(3x+2)]}{[(3x+2)^2]^2}$

$\displaystyle =\frac{[(3x+2)^2(2)]-[(2x-5)(9x^2+6x+6x+4)]}{[3x+2)^2]^2}$

$\displaystyle =\frac{[(3x+2)^2(2)]-[(2x-5)(9x^2+12x)]}{[(3x+2)^2]^2}$

$\displaystyle =\frac{[(3x+2)^2(2)]-[(2x-5)(18x+12)]}{[(3x+2)^2]^2}$

$\displaystyle =\frac{[((3x+2)(3x+2)(2)]-(36x+24x-90x-60)}{[(3x+2)(3x+2)(3x+2)(3x+2)]}$

$\displaystyle =\frac{((3x+2)(3x+2)(2)-36x-24x+90x+60} {(3x+2)(3x+2)(3x+2)(3x+2)}$

$\displaystyle =\frac{(9x^2+12x+4)(2)+30x+60}{54x+78x+24}$

$\displaystyle =\frac{(18x^2+24x+8)+30x+60}{132x+24}$

$\displaystyle =\frac{18x^2+54x+68}{132x+24}$

ok i know thats alot to read but if anyone is able to let me know if ive gone wrong anywhere it would be much appreciated.