Answer:
[tex]\frac{y}{x} + \frac{2y}{x+4} =3[/tex]
[tex]\frac{y}{x} + \frac{2y}{x+4} = \frac{2yx+yx+4y}{x^2+4x} = \frac{3yx+4y}{x^2+4x}[/tex]
So now we have (3yx+4y)/(x^2+4x) = 3
multiply both sides by (x^2+4x)
u get
[tex]3yx+4y = 3(x^2+4x) ............Next........... 3yx+4y = 3x^2+12x[/tex]
[tex]3x+4 =\frac{3x^2+12x}{y}[/tex]
divide by 3x+4 and multiply by y to get...
3= [tex]\frac{3x^2+12x}{3x+4}[/tex]
