(a)[2]
Show that the fraction of the balls in bag $A$ that are red is $\frac{x-6}{2x}$.
(b)[2]
Bag $B$ also has red balls and green balls. The number of red balls in bag $B$ is $x$. The number of green balls in bag $B$ is $4$ times the number of green balls in bag $A$. Show that the fraction of the balls in bag $B$ that are red is $\frac{x}{3x+12}$.
(c)[3]
From $\frac{x-6}{2x} = \frac{x}{3x+12}$, Show that $x^2 - 6x - 72 = 0$.
(d)[2]
Solve by factorisation the equation $x^2 - 6x - 72 = 0$.
(e)[1]
$x$ gives the total number of balls in bag $A$. Use your answer to part (d) to find how many green balls are in bag $A$.