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Maths Question 3 – JEE-MAIN 2026

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Let α,β∈R be such that the system of linear equations x+2y+z=5 2x+y+αz=5 8x+4y+βz=18 has no solution. Then βα is equal to :

A system of linear equations has no solution if, after algebraic manipulation, it leads to a contradictory statement like 0=k where k≠0.

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Ninja Strategy•Algebraic Manipulation for Contradiction

By observing the relationship between the second and third equations, one can substitute and simplify to directly find the condition for a contradiction (0=k,k≠0), which implies no solution.

Step 1: Identify relationships between equations✦ Active

The given system of equations is: (1) x+2y+z=5 (2) 2x+y+αz=5 (3) 8x+4y+βz=18 Notice that the coefficients of x and y in equation (3) are four times the coefficients of x and y in equation (2). We can rewrite equation (3) as 4(2x+y)+βz=18.

Step 2: Substitute and derive the condition for no solution○ Expand

From equation (2), we have 2x+y=5−αz. Substitute this into the modified equation (3):

4(5−αz)+βz=18 20−4αz+βz=18 (β−4α)z=18−20 (β−4α)z=−2

For the system to have no solution, this equation must be a contradiction. This occurs if the coefficient of z is zero, but the right-hand side is non-zero. Thus, we must have β−4α=0 and −2≠0. The second part is true. So, the condition for no solution is β−4α=0⟹β=4α.

💡 Teacher's Secret Hint

Remember that for no solution, the coefficient of the variable must be zero while the constant term is non-zero.

Step 3: Calculate the required ratio○ Expand

Given β=4α, and assuming α≠0 (as βα is asked), the ratio is:

βα=4
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