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 [2026]

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 [2026]

Smart Achievers · Accepted Answer

(2)

$△ = | \begin{matrix} 1 & 5 & 6 \\ 2 & 3 & 4 \\ 1 & 6 & a \end{matrix} | = 0$

$\Rightarrow (3 a - 24) - 5 (2 a - 4) + 6 (9) = 0$

$\Rightarrow - 7 a + 50 = 0 \Rightarrow a = \frac{50}{7}$

Now, $△_{z} = = 0 \Rightarrow | \begin{matrix} 1 & 5 & 4 \\ 2 & 3 & 7 \\ 1 & 6 & b \end{matrix} | = 0$

$\Rightarrow (3 b - 42) - 5 (2 b - 7) + 4 (9) = 0$

$\Rightarrow - 7 b + 29 = 0 \Rightarrow b = \frac{29}{7}$

Now, a – b = 3

$∴$ x – y = 3

Solution

Q.
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 [2026]

1 y – x = 3

2 x – y = 3

3 x + y = 11

4 x + y = 12

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