JEE Main 2026 — Mathematics Algebra

Given the equation:

$$50\left(\dfrac{2x}{1+3i} - \dfrac{y}{1-2i}\right) = 31 + 17i$$50(1+3i2x​−1−2iy​)=31+17i

Rationalizing the denominators inside the parentheses:

$$\dfrac{2x}{1+3i} = \dfrac{2x(1-3i)}{(1+3i)(1-3i)} = \dfrac{2x(1-3i)}{1^2 + 3^2} = \dfrac{2x(1-3i)}{10} = \dfrac{x(1-3i)}{5}$$1+3i2x​=(1+3i)(1−3i)2x(1−3i)​=12+322x(1−3i)​=102x(1−3i)​=5x(1−3i)​

$$\dfrac{y}{1-2i} = \dfrac{y(1+2i)}{(1-2i)(1+2i)} = \dfrac{y(1+2i)}{1^2 + 2^2} = \dfrac{y(1+2i)}{5}$$1−2iy​=(1−2i)(1+2i)y(1+2i)​=12+22y(1+2i)​=5y(1+2i)​

Substituting these back into the given equation:

$$50\left(\dfrac{x(1-3i)}{5} - \dfrac{y(1+2i)}{5}\right) = 31 + 17i$$50(5x(1−3i)​−5y(1+2i)​)=31+17i

$$10(x(1-3i) - y(1+2i)) = 31 + 17i$$10(x(1−3i)−y(1+2i))=31+17i

$$10(x - y) - 10i(3x + 2y) = 31 + 17i$$10(x−y)−10i(3x+2y)=31+17i

Equating the real and imaginary parts on both sides:

$$10(x - y) = 31 \Rightarrow 10x - 10y = 31$$10(x−y)=31⇒10x−10y=31

$$-10(3x + 2y) = 17 \Rightarrow -30x - 20y = 17$$−10(3x+2y)=17⇒−30x−20y=17

Multiplying the first equation by $3$3 gives:

$$30x - 30y = 93$$30x−30y=93

Adding this to the second equation:

$$(-30x - 20y) + (30x - 30y) = 17 + 93$$(−30x−20y)+(30x−30y)=17+93

$$-50y = 110 \Rightarrow y = -\dfrac{11}{5}$$−50y=110⇒y=−511​

Substituting $y = -\dfrac{11}{5}$y=−511​ into $10x - 10y = 31$10x−10y=31:

$$10x - 10\left(-\dfrac{11}{5}\right) = 31$$10x−10(−511​)=31

$$10x + 22 = 31 \Rightarrow 10x = 9 \Rightarrow x = \dfrac{9}{10}$$10x+22=31⇒10x=9⇒x=109​

Now, finding the value of $10(x - 3y)$10(x−3y):

$$10(x - 3y) = 10x - 30y = 10\left(\dfrac{9}{10}\right) - 30\left(-\dfrac{11}{5}\right)$$10(x−3y)=10x−30y=10(109​)−30(−511​)

$$10(x - 3y) = 9 + 66 = 75$$10(x−3y)=9+66=75

Answer: $75$75