Question

If the solution of the system of simultaneous equations \[ \frac{1}{x}+\frac{2}{y}-\frac{3}{z}-1=0, \] \[ \frac{2}{x}-\frac{4}{y}+\frac{3}{z}-1=0 \] and \[ \frac{3}{x}+\frac{6}{y}-\frac{6}{z}-4=0 \] is \(x=\alpha,\ y=\beta,\ z=\gamma\), then \(\alpha^2+\gamma^2=\)

• A. \(5\beta\)
• B. \(\beta^2\)
• C. \(3\beta\)
• D. \(2\beta^2\)

Hint:

Whenever equations contain terms like \[ \frac{1}{x},\ \frac{1}{y},\ \frac{1}{z}, \] replace them with new variables to convert the system into linear equations.

Answer 1

The Correct Option is A

Step 1: Substitute to convert to linear equations.
Let $a = \frac{1}{x}$, $b = \frac{1}{y}$, $c = \frac{1}{z}$. The system becomes: \[ a + 2b - 3c = 1, \quad 2a - 4b + 3c = 1, \quad 3a + 6b - 6c = 4 \]

Step 2: Solve for $c$ using equations 1 and 3.
Triple equation 1: $3a + 6b - 9c = 3$. Subtract from equation 3: $(3a+6b-6c) - (3a+6b-9c) = 4-3 \Rightarrow 3c = 1 \Rightarrow c = \frac{1}{3}$.

Step 3: Solve for $b$.
Add equations 1 and 2: $(a+2b-3c) + (2a-4b+3c) = 1+1 \Rightarrow 3a - 2b = 2$. Also from eq 1: $a = 1 - 2b + 3(\frac{1}{3}) = 2 - 2b$. Substitute: $3(2-2b) - 2b = 2 \Rightarrow 6-6b-2b=2 \Rightarrow 8b=4 \Rightarrow b=\frac{1}{2}$.

Step 4: Solve for $a$.
$a = 2 - 2(\frac{1}{2}) = 2 - 1 = 1$.

Step 5: Find $x, y, z$.
$a=1 \Rightarrow x=1$, $b=\frac{1}{2} \Rightarrow y=2$, $c=\frac{1}{3} \Rightarrow z=3$. So $\alpha=1$, $\beta=2$, $\gamma=3$.

Step 6: Compute the required expression.
\[ \alpha^2 + \gamma^2 = 1 + 9 = 10 = 5(2) = 5\beta \] \[ \boxed{5\beta} \]