QJEEM25Ma5771: JEE Main 2025 Mathematics - Class 12 JEEM25Ma5771 Solution: If the system of linear equations : where , has infinitely many …

We begin with the system:

$\begin{aligned} x+y+2z &= 6, \\ 2x+3y+az &= a+1, \\ -x-3y+bz &= 2b. \end{aligned}$

Step 1. Solve the first equation for $x$ :

$x = 6 - y - 2z.$

Step 2. Substitute $x = 6-y-2z$ into the second equation:

$2(6-y-2z) + 3y + az = a+1.$

Expanding and simplifying:

$12 - 2y - 4z + 3y + az = a+1 \quad \Longrightarrow \quad y + (a-4)z = a - 11.$

Call this Equation (I).

Step 3. Substitute $x = 6-y-2z$ into the third equation:

$-(6-y-2z) - 3y + bz = 2b.$

Expanding and simplifying:

$-6 + y + 2z - 3y + bz = 2b \quad \Longrightarrow \quad -2y + (b+2)z = 2b + 6.$

Call this Equation (II).

Step 4. For the system to have infinitely many solutions, the two equations in $y$ and $z$ must be dependent—that is, one must be a constant multiple of the other. Assume there exists a constant $k$ such that

$-2 = k \cdot 1 \quad \Longrightarrow \quad k = -2.$

Apply this to the coefficient of $z$ and the constant term.

For the $z$ -coefficient in Equations (I) and (II):

$b+2 = k(a-4) = -2(a-4) = -2a+8.$

Thus,

$b = -2a+6.$

For the constant term:

$2b+6 = k(a-11) = -2(a-11) = -2a + 22.$

Substitute $b = -2a+6$ into this equation:

$2(-2a+6) + 6 = -2a + 22 \quad \Longrightarrow \quad -4a + 12 + 6 = -2a + 22.$

Simplify:

$-4a + 18 = -2a + 22.$

Solve for $a$ :

$-4a + 18 + 4a = -2a + 22 + 4a \quad \Longrightarrow \quad 18 = 2a + 22,$

$2a = 18 - 22 = -4 \quad \Longrightarrow \quad a = -2.$

Substitute $a = -2$ into $b = -2a+6$ :

$b = -2(-2) + 6 = 4 + 6 = 10.$

Step 5. We now compute

$7a + 3b = 7(-2) + 3(10) = -14 + 30 = 16.$

Thus, the value of $7a+3b$ is

$\boxed{16}.$