Question

If the system of equations \( x + 5y + 6z = 4 \), \( 2x + 2y + 4z = 1 \) and \( x + y + az = b \) has infinite numbers of solutions then point \( (a,b) \) lies on-

• A. \( x - 2y = 1 \)
• B. \( x - y = 3 \)
• C. \( x + y = 2 \)
• D. \( y - x = 3 \)

Hint:

For infinitely many solutions in a system of three linear equations, one equation must be obtainable as a linear combination of the other two. Compare coefficients carefully to find the required parameters.

Answer 1

The Correct Option is A

Step 1: Write the three equations.
Given equations are
\[ x + 5y + 6z = 4 \] \[ 2x + 2y + 4z = 1 \] \[ x + y + az = b \] For the system to have infinitely many solutions, the third equation must be a linear combination of the first two equations.

Step 2: Express the third equation as a combination of the first two.
Let \[ \lambda(x+5y+6z=4)+\mu(2x+2y+4z=1) \] give the third equation \[ x+y+az=b. \] Comparing coefficients, we get
\[ \lambda+2\mu=1 \] \[ 5\lambda+2\mu=1 \] Subtracting the first equation from the second,
\[ 4\lambda=0 \] \[ \lambda=0. \] Then from \[ \lambda+2\mu=1, \] we get \[ 2\mu=1 \Rightarrow \mu=\frac{1}{2}. \]
Step 3: Find \( a \) and \( b \).
Now compare the coefficient of \( z \):
\[ a=6\lambda+4\mu=6(0)+4\left(\frac{1}{2}\right)=2. \] Compare constants:
\[ b=4\lambda+\mu=4(0)+\frac{1}{2}=\frac{1}{2}. \] So the point is
\[ (a,b)=\left(2,\frac{1}{2}\right). \]
Step 4: Check which option is satisfied by \( (a,b) \).
Substitute \( x=2 \) and \( y=\frac{1}{2} \) into the options.
For option (A):
\[ x-2y=2-2\left(\frac{1}{2}\right)=2-1=1 \] which is true.
Hence, the point \( (a,b) \) lies on
\[ x-2y=1. \] Final Answer: \( x-2y=1 \)