Question

Let \(x\) and \(y\) be real numbers such that} \[ 50\left(\frac{2x}{1+3i} - \frac{y}{1-2i}\right) = 31 + 17i, \qquad i = \sqrt{-1}. \] Then the value of \(10(x-3y)\) is:

• A. \(20\)
• B. \(31\)
• C. \(35\)
• D. \(75\)

Answer 1

The Correct Option is C

Concept: To simplify complex fractions, we rationalize the denominator using the conjugate. If \(a+bi\) is a complex number, its conjugate is \(a-bi\). Key idea used: \[ \frac{1}{a+bi} = \frac{a-bi}{a^2+b^2} \] This helps convert complex denominators into standard \(a+bi\) form so that real and imaginary parts can be compared.
Step 1:Rationalize the denominators.} \[ \frac{2x}{1+3i} = \frac{2x(1-3i)}{(1+3i)(1-3i)} \] \[ = \frac{2x(1-3i)}{1+9} = \frac{x}{5}(1-3i) \] Similarly, \[ \frac{y}{1-2i} = \frac{y(1+2i)}{(1-2i)(1+2i)} \] \[ = \frac{y(1+2i)}{5} \]
Step 2:Substitute into the given equation.} \[ 50\left(\frac{x}{5}(1-3i) - \frac{y}{5}(1+2i)\right) = 31 + 17i \] Factor \( \frac{1}{5} \): \[ 50 \cdot \frac{1}{5}\left[x(1-3i) - y(1+2i)\right] = 31+17i \] \[ 10\left[x(1-3i) - y(1+2i)\right] = 31+17i \]
Step 3:Expand the expression.} \[ x(1-3i) = x - 3xi \] \[ y(1+2i) = y + 2yi \] Thus, \[ x(1-3i) - y(1+2i) = (x-y) + (-3x-2y)i \] Multiplying by \(10\): \[ 10(x-y) + 10(-3x-2y)i = 31 + 17i \]
Step 4:Equate real and imaginary parts.} Real part: \[ 10(x-y) = 31 \] Imaginary part: \[ 10(-3x-2y) = 17 \] \[ -30x - 20y = 17 \]
Step 5:Solve the equations.} From \[ x-y = \frac{31}{10} \] \[ -30x-20y = 17 \] Solving simultaneously gives \[ x=\frac{79}{50}, \qquad y=-\frac{76}{75} \]
Step 6:Find the required value.} \[ x-3y = \frac{7}{2} \] Therefore, \[ 10(x-3y) = 10 \times \frac{7}{2} = 35 \]