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Maths Question 2 – JEE-MAIN 2026

Let x and y be real numbers such that 50(2x1+3iy12i)=31+17i, i=1. Then the value of 10(x3y) is :

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

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(13i)(1+3i)(13i)=2x(13i)12+32=2x(13i)10=x53x5i
y12i=y(1+2i)(12i)(1+2i)=y(1+2i)12+(2)2=y(1+2i)5=y5+2y5i

Substitute these back into the given equation:

50((x53x5i)(y5+2y5i))=31+17i
50(xy53x+2y5i)=31+17i
10(xy)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(xy)=31(Equation 1)
10(3x+2y)=1710(3x+2y)=17(Equation 2)

From Equation 1: xy=3110

From Equation 2: 3x+2y=1710

Multiply the first equation by 2: 2x2y=6210=315

Add this to the second equation: (2x2y)+(3x+2y)=3151710

5x=621710=4510=92

So, x=910.

Substitute x back into xy=3110:

910y=3110y=9103110=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(x3y) using the obtained values of x and y:

10(x3y)=10(9103(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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