Question

If the system of equations} \[ x + 5y + 6z = 4 \] \[ 2x + 3y + 4z = 7 \] \[ x + 6y + az = b \] has infinitely many solutions, then the point \( (a, b) \) lies on the line}

• A. \( y - x = 3 \)
• B. \( x - y = 3 \)
• C. \( x + y = 11 \)
• D. \( x + y = 12 \)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A system of three linear equations has infinitely many solutions if the determinant of the coefficient matrix is zero and the equations are linearly dependent.
Alternatively, we can eliminate one variable and force the resulting two equations in two variables to be identical.
Step 2: Key Formula or Approach:
We will eliminate \(x\) from the given system to form two identical equations in \(y\) and \(z\).
Equation 1 is \(x + 5y + 6z = 4\).
Equation 2 is \(2x + 3y + 4z = 7\).
Equation 3 is \(x + 6y + az = b\).
Step 3: Detailed Explanation:
Perform the operation (Equation 2) - \(2 \times\) (Equation 1) to eliminate \(x\).
\[ (2x + 3y + 4z) - 2(x + 5y + 6z) = 7 - 2(4) \] \[ -7y - 8z = -1 \implies 7y + 8z = 1 \] This can be written as \(y + \frac{8}{7}z = \frac{1}{7}\).
Now, perform (Equation 3) - (Equation 1) to eliminate \(x\) again.
\[ (x + 6y + az) - (x + 5y + 6z) = b - 4 \] \[ y + (a - 6)z = b - 4 \] For the system to have infinitely many solutions, these two newly formed equations must represent the same line.
By comparing the coefficients of \(z\) and the constant terms, we get the values of \(a\) and \(b\).
\[ a - 6 = \frac{8}{7} \implies a = 6 + \frac{8}{7} = \frac{50}{7} \] \[ b - 4 = \frac{1}{7} \implies b = 4 + \frac{1}{7} = \frac{29}{7} \] Now we have the coordinates of the point \((a, b) = \left(\frac{50}{7}, \frac{29}{7}\right)\).
We check which option satisfies this point by finding \(a - b\).
\[ a - b = \frac{50}{7} - \frac{29}{7} = \frac{21}{7} = 3 \] Thus, the point satisfies the line equation \(x - y = 3\).
Step 4: Final Answer:
The point \((a, b)\) lies on the line \(x - y = 3\).