The simplified answer is [tex]\frac{\left(12 x^{2}+7 x y-4 y z-3 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}[/tex].
Step-by-step explanation:
[tex]$\frac{3 y+2 x}{z+2 x}-\frac{2 y-3 x}{3 x+y}-\frac{2 z(y+3 x)}{6 x^{2}+3 x z+2 x y+y z}[/tex]
To add or subtract denominators of the fraction must be same.
If it is not the same, we must take LCM of the denominators. and so we can add the fractions.
To make the denominator same multiply the 1st term [tex](\frac{3x+y}{3x+y})[/tex] and 2nd term by [tex](\frac{z+2x}{z+2x})[/tex]
= [tex]\frac{(3 y+2 x)(3 x+y)}{(z+2 x)(3 x+y)}-\frac{(2 y-3 x)(z+2 x)}{(3 x+y)(z+2 x)}-\frac{2 z(y+3 x)}{6 x^{2}+3 x z+2 x y+y z}[/tex]
LCM of the denominators is 6x²+ 3xz + 2xy +yz.
Multiply the factors in the numerator.
= [tex]\frac{\left(6 x^{2}+3 y^{2}+11 x y\right)}{(z+2 x)(3 x+y)}-\frac{\left(2 y z+4 x y-3 x z-6 x^{2}\right)}{(3 x+y)(z+2 x)}-\frac{2 z y+6 x z}{6 x^{2}+3 x z+2 x y+y z}[/tex]
Now, the denominators are same, you can subtract it.
= [tex]\frac{\left(6 x^{2}+6 x^{2}+11 x y-4 x y-2 y z-2 y z+3 x z-6 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}[/tex]
= [tex]\frac{\left(12 x^{2}+7 x y-4 y z-3 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}[/tex]
Thus the simplified solution is [tex]\frac{\left(12 x^{2}+7 x y-4 y z-3 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}[/tex]
