Maths Q5 – JEE-MAIN 2025

Let the system of equations

x + 5 y - z = 1

4 x + 3 y - 3 z = 7

24 x + y + λ z = μ

λ, μ \in R

, have infinitely many solutions. Then the number of the solutions of this system, if

x, y, z

are integers and satisfy

7 \leq x + y + z \leq 77

, is :

Clue 1 of 3: Condition for Infinitely Many Solutions

For a system of linear equations $A X = B$ to have infinitely many solutions, the rank of the coefficient matrix $A$ must be equal to the rank of the augmented matrix $[A | B]$ , and this rank must be less than the number of variables.

🧠 Full Solution Path

Step 1: Determine values of

λ

and

μ

for infinitely many solutions✦ Active

For the system to have infinitely many solutions, the third equation must be a linear combination of the first two. Let $E_{3} = a E_{1} + b E_{2}$ :

a (x + 5 y - z) + b (4 x + 3 y - 3 z) = 24 x + y + λ z

⟹ (a + 4 b) x + (5 a + 3 b) y + (- a - 3 b) z = 24 x + y + λ z

Comparing coefficients and constant terms:

a + 4 b = 24 (1)

5 a + 3 b = 1 (2)

- a - 3 b = λ (3)

a + 7 b = μ (4)

Solving (1) and (2): Multiply (1) by 5 and (2) by 1, then subtract: $5 (a + 4 b) - (5 a + 3 b) = 5 (24) - 1 ⟹ 17 b = 119 ⟹ b = 7$ . Substitute $b = 7$ into (1): $a + 4 (7) = 24 ⟹ a + 28 = 24 ⟹ a = - 4$ . Now, substitute $a = - 4$ and $b = 7$ into (3) and (4):

λ = - (- 4) - 3 (7) = 4 - 21 = - 17

μ = - 4 + 7 (7) = - 4 + 49 = 45

Step 2: Find the general solution

(x, y, z)

in terms of an integer parameter○ Expand

Since the system has infinitely many solutions, we only need to solve the first two equations:

x + 5 y - z = 1 (E_{1})

4 x + 3 y - 3 z = 7 (E_{2})

Multiply $(E_{1})$ by 3: $3 x + 15 y - 3 z = 3$ . Subtract this from $(E_{2})$ :

(4 x + 3 y - 3 z) - (3 x + 15 y - 3 z) = 7 - 3

x - 12 y = 4 ⟹ x = 12 y + 4

Substitute $x = 12 y + 4$ into $(E_{1})$ :

(12 y + 4) + 5 y - z = 1

17 y + 4 - z = 1 ⟹ z = 17 y + 3

Let $y = k$ , where $k$ is an integer. Then the general solution is:

x = 12 k + 4

y = k

z = 17 k + 3

💡 Teacher's Secret Hint

Remember that for infinitely many solutions, the system is consistent and dependent, allowing one or more variables to be expressed in terms of parameters.

Step 3: Apply the inequality condition and count the solutions○ Expand

We are given that $x, y, z$ are integers and satisfy $7 \leq x + y + z \leq 77$ . Calculate the sum $x + y + z$ in terms of $k$ :

x + y + z = (12 k + 4) + k + (17 k + 3) = 30 k + 7

Now apply the inequality:

7 \leq 30 k + 7 \leq 77

Subtract 7 from all parts of the inequality:

0 \leq 30 k \leq 70

Divide by 30:

0 \leq k \leq \frac{70}{30}

0 \leq k \leq \frac{7}{3}

0 \leq k \leq 2.333 . . .

Since $k$ must be an integer, the possible values for $k$ are $0, 1, 2$ . Each value of $k$ corresponds to a unique integer solution $(x, y, z)$ . Therefore, there are 3 solutions.

💡 Teacher's Secret Hint

Ensure to consider only integer values for the parameter $k$ as $x, y, z$ are integers.

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Maths Question 5 – JEE-MAIN 2025

Clue 1 of 3: Condition for Infinitely Many Solutions

🧠 Full Solution Path