Question

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:

• A. \(18\)
• B. \(20\)
• C. \(15\)
• D. \(14\)

Hint:

When several linear expressions repeat with the same structure (like $3b - 3c$ here), use substitution to reduce the system to two variables. This often simplifies the equations dramatically.

Answer 1

The Correct Option is A

Step 1: Since there are 3 unknowns but only 2 equations, fix \(c = t\) as a free parameter and solve for \(a\) and \(b\) in terms of \(t\). From \(a - 6b + 6c = 4\): \(a = 4 + 6b - 6t\). Substitute into \(6a + 3b - 3c = 50\): \(6(4 + 6b - 6t) + 3b - 3t = 50 \Rightarrow 24 + 36b - 36t + 3b - 3t = 50 \Rightarrow 39b = 26 + 39t \Rightarrow b = \frac{2}{3} + t\).
Step 2: Then \(a = 4 + 6\left(\frac{2}{3} + t\right) - 6t = 4 + 4 + 6t - 6t = 8\). Notice \(a\) comes out to \(8\) no matter what \(t\) is.
Step 3: Substitute into the target expression: \(2a + 3b - 3c = 2(8) + 3\left(\frac{2}{3} + t\right) - 3t = 16 + 2 + 3t - 3t = 18\). The parameter \(t\) cancels out completely, confirming the answer does not depend on which valid \((a,b,c)\) triple you pick.
\[ \boxed{2a + 3b - 3c = 18} \]