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:

Notice that $-6b+6c$ is just $-2$ times the block $(3b-3c)$, which also appears in the second equation and in the expression you need to find. Instead of solving for $a$, $b$, and $c$ separately, try to write the target expression as a combination of the two given equations.

Answer 1

The Correct Option is A

Approach: Don't chase all three letters. Both equations and the target share the block \((3b - 3c)\), so treat that block as a single unknown and you are left with just two equations in two quantities.

Step 1: Bundle the repeated block.
Let \(X = 3b - 3c\). Notice \(-6b + 6c = -2(3b - 3c) = -2X\).
First equation: \(a - 6b + 6c = 4\) becomes
\[ a - 2X = 4. \tag{i} \]
Second equation: \(6a + 3b - 3c = 50\) becomes
\[ 6a + X = 50. \tag{ii} \]
Target: \(2a + 3b - 3c = 2a + X\).

Step 2: Solve (i) and (ii).
From (ii): \(X = 50 - 6a\). Put into (i):
\[ a - 2(50 - 6a) = 4 \Rightarrow a - 100 + 12a = 4 \Rightarrow 13a = 104 \Rightarrow a = 8. \]
Then \(X = 50 - 6(8) = 2\).

Step 3: Evaluate the target.
\[ 2a + X = 2(8) + 2 = 18. \]

\(2a + 3b - 3c = 18\) — option (1).

Answer 2

Approach: Rewrite both equations in terms of \(a\) and the block \(u = b - c\) (since \(-6b + 6c = -6u\) and \(3b - 3c = 3u\)), turning this into a clean two-variable linear system that can be solved directly by Cramer's Rule, without any back-substitution.

Step 1: Set up the system.
\[ a - 6u = 4, \qquad 6a + 3u = 50. \]

Step 2: Apply Cramer's Rule.
The coefficient determinant is \( D = (1)(3) - (-6)(6) = 3 + 36 = 39 \).
Swapping in the constants for the \(a\)-column gives \( D_a = (4)(3) - (-6)(50) = 12 + 300 = 312 \), so \( a = \frac{312}{39} = 8 \).
Swapping in the constants for the \(u\)-column gives \( D_u = (1)(50) - (4)(6) = 50 - 24 = 26 \), so \( u = \frac{26}{39} = \frac{2}{3} \).

Step 3: Evaluate the target.
\[ 2a + 3b - 3c = 2a + 3u = 2(8) + 3\left(\frac{2}{3}\right) = 16 + 2 = 18. \]

Value = 18, option (1).

Answer 3

Concept:

• Instead of solving for a, b, c individually, look for constants that let you combine the two given equations directly into the target expression.
• Since the equations are linear, the expression $2a+3b-3c$ can be written as a weighted sum $k_1(\text{Eq 1}) + k_2(\text{Eq 2})$ for suitable constants $k_1, k_2$.
• Once those constants are found, the answer follows from the right-hand sides of the equations alone, with no need to solve for $a$, $b$, or $c$ separately.

Step 1: Set up the target as a combination.
We want $2a+3b-3c = k_1(a-6b+6c) + k_2(6a+3b-3c)$ for some numbers $k_1,k_2$.
Matching the coefficient of $a$: $k_1+6k_2=2$.
Matching the coefficient of the block $3b-3c$: $-6k_1+3k_2=3$.

Step 2: Solve for $k_1$ and $k_2$.
From the second equation: $-2k_1+k_2=1$, so $k_2=1+2k_1$.
Substitute into the first: $k_1+6(1+2k_1)=2 \Rightarrow 13k_1=-4 \Rightarrow k_1=-\dfrac{4}{13}$.
Then $k_2=1+2\left(-\dfrac{4}{13}\right)=\dfrac{5}{13}$.

Step 3: Apply the combination to the two given right-hand sides.
Since $a-6b+6c=4$ and $6a+3b-3c=50$:
$2a+3b-3c = k_1(4)+k_2(50) = -\dfrac{4}{13}(4)+\dfrac{5}{13}(50) = -\dfrac{16}{13}+\dfrac{250}{13} = \dfrac{234}{13}$.

Step 4: Simplify.
$\dfrac{234}{13}=18$.

Final Answer: $2a+3b-3c=18$