If $$a-6b+6c=4$$ and $$6a+3b-3c=50$$, where a, b and c are real numbers, the value of $$2a+3b-3c$$ is

Given, $$a-6b+6c=4$$ --->(1)

and, $$6a+3b-3c=50$$ ---->(2)

Multiplying eqn (1) with $$x$$, $$ax-6bx+6cx=4x$$

And multiplying eqn (2) with $$y$$, $$6ay+3by-3cy=50y$$

So, $$x+6y=2$$ and $$3y-6x=3$$

So, $$x+6y=2$$ ---->(3) and $$-2x+y=1$$ ---->(4)

Multiplying eqn (4) with 6 and subtracting from eqn (3),

$$13x=-4$$

So, $$x=-\dfrac{4}{13}$$

Putting the value of $$x$$ in equation (4),

$$y=\dfrac{5}{13}$$

So, the final answer is $$4x+5y$$

=$$4\left(-\frac{4}{13}\right)+50\left(\frac{5}{13}\right)=-\frac{16}{13}+\frac{250}{13}=\frac{234}{13}=18$$

So, correct answer is $$18$$.