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: Introduce substitutions.
Let \[ a=\frac{1}{x}, \quad b=\frac{1}{y}, \quad c=\frac{1}{z} \] Then the equations become \[ a+2b-3c=1 \] \[ 2a-4b+3c=1 \] \[ 3a+6b-6c=4 \]

Step 2: Solve the linear system.
From the first equation, \[ a=1-2b+3c \] Substitute into the second equation: \[ 2(1-2b+3c)-4b+3c=1 \] \[ 2-4b+6c-4b+3c=1 \] \[ -8b+9c=-1 \] \[ 8b-9c=1 \] Now substitute \(a=1-2b+3c\) into the third equation: \[ 3(1-2b+3c)+6b-6c=4 \] \[ 3-6b+9c+6b-6c=4 \] \[ 3+3c=4 \] \[ 3c=1 \] \[ c=\frac{1}{3} \] Substituting into \[ 8b-9c=1, \] we get \[ 8b-3=1 \] \[ 8b=4 \] \[ b=\frac{1}{2} \] Now, \[ a=1-2\left(\frac{1}{2}\right)+3\left(\frac{1}{3}\right) \] \[ a=1-1+1 \] \[ a=1 \]

Step 3: Find \(x,y,z\).
Since \[ a=\frac{1}{x}=1, \] we get \[ x=1 \] Since \[ b=\frac{1}{y}=\frac{1}{2}, \] we get \[ y=2 \] Since \[ c=\frac{1}{z}=\frac{1}{3}, \] we get \[ z=3 \] Thus, \[ \alpha=1,\quad \beta=2,\quad \gamma=3 \]

Step 4: Compute \(\alpha^2+\gamma^2\).
\[ \alpha^2+\gamma^2=1^2+3^2 \] \[ =1+9 \] \[ =10 \] Also, \[ 5\beta=5(2)=10 \] Hence, \[ \alpha^2+\gamma^2=5\beta \]

Step 5: Final conclusion.
Therefore, \[ \boxed{5\beta} \]