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Maths Question 2 – JEE-MAIN 2026

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Let x and y be real numbers such that 50(2x1+3i−y1−2i)=31+17i, i=−1. Then the value of 10(x−3y) is :

To simplify complex fractions like a+bic+di, multiply the numerator and denominator by the conjugate of the denominator, which is c−di.

● Video Walkthrough
Step 1: Simplify the Complex Expression✦ Active

First, simplify each complex fraction by multiplying the numerator and denominator by the conjugate of the denominator:

2x1+3i=2x(1−3i)(1+3i)(1−3i)=2x(1−3i)12+32=2x(1−3i)10=x5−3x5i
y1−2i=y(1+2i)(1−2i)(1+2i)=y(1+2i)12+(−2)2=y(1+2i)5=y5+2y5i

Substitute these back into the given equation:

50((x5−3x5i)−(y5+2y5i))=31+17i
50(x−y5−3x+2y5i)=31+17i
10(x−y)−10(3x+2y)i=31+17i
Step 2: Equate Real and Imaginary Parts○ Expand

By equating the real and imaginary parts of the equation, we get a system of two linear equations:

10(x−y)=31(Equation 1)
−10(3x+2y)=17⟹10(3x+2y)=−17(Equation 2)

From Equation 1: x−y=3110

From Equation 2: 3x+2y=−1710

Multiply the first equation by 2: 2x−2y=6210=315

Add this to the second equation: (2x−2y)+(3x+2y)=315−1710

5x=62−1710=4510=92

So, x=910.

Substitute x back into x−y=3110:

910−y=3110⟹y=910−3110=−2210=−115
💡 Teacher's Secret Hint

Ensure careful algebraic manipulation when solving the system of equations to avoid sign errors.

Step 3: Calculate the Required Value○ Expand

Now, calculate the value of 10(x−3y) using the obtained values of x and y:

10(x−3y)=10(910−3(−115))
=10(910+335)
=10(910+6610)
=10(9+6610)
=10(7510)
=75
💡 Teacher's Secret Hint

Double-check your arithmetic, especially when dealing with fractions and negative signs.

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