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

The correct option is 18.

We are given the following two equations:

$a-6b+6c=4$

$6a+3b-3c=50$

We want to find the value of:

$2a+3b-3c$

First, observe that in all these expressions, the variables $b$ and $c$ appear together with proportional coefficients. We can factor out common terms to simplify the system.

From the first equation, we can factor out $-6$ from the $b$ and $c$ terms:

$a-6(b-c)=4$

From the second equation, we can factor out $3$:

$6a+3(b-c)=50$

Notice that the expression $(b-c)$ appears in both equations. To make solving easier, let's substitute a new variable, say $x=b-c$. Our equations become:

Equation 1: $a-6x=4$

Equation 2: $6a+3x=50$

Now we have a simple system of two linear equations with two variables. We can solve for $a$ and $x$. From Equation 1, express $a$ in terms of $x$:

$a=6x+4$

Substitute this expression for $a$ into Equation 2:

$6(6x+4)+3x=50$

Distribute the 6:

$36x+24+3x=50$

Combine like terms:

$39x+24=50$

Subtract 24 from both sides:

$39x=26$

Solve for $x$:

$x=\frac{26}{39}$

Both 26 and 39 are divisible by 13, so simplify the fraction:

$x=\frac{2}{3}$

Now that we have $x$, we can find the value of $a$ using our earlier expression $a=6x+4$:

$a=6\left(\frac{2}{3}\right)+4$

$a=4+4=8$

Finally, we need to evaluate the target expression: $2a+3b-3c$. We can rewrite this by factoring out the 3 from the $b$ and $c$ terms:

$2a+3(b-c)$

Substitute back our variable $x=b-c$:

$2a+3x$

Substitute the known values of $a=8$ and $x=\frac{2}{3}$:

$2\left(8\right)+3\left(\frac{2}{3}\right)$

$16+2=18$

Therefore, the value of the expression is 18.