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

Step 1: Understanding the Concept:
To solve an equation involving complex numbers, we simplify each term to the form \(a + bi\).
Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal.
Step 2: Key Formula or Approach:
1. Rationalize denominators: Multiply the numerator and denominator by the conjugate of the denominator.
2. Compare Real Parts and Imaginary Parts.
Step 3: Detailed Explanation:
Rationalize the fractions:
\[ \frac{2x}{1+3i} = \frac{2x(1-3i)}{(1+3i)(1-3i)} = \frac{2x(1-3i)}{1^2 + 3^2} = \frac{2x(1-3i)}{10} = \frac{x(1-3i)}{5} \]
\[ \frac{y}{1-2i} = \frac{y(1+2i)}{(1-2i)(1+2i)} = \frac{y(1+2i)}{1^2 + 2^2} = \frac{y(1+2i)}{5} \]
Substitute back into the equation:
\[ 50 \left( \frac{x(1-3i)}{5} - \frac{y(1+2i)}{5} \right) = 31 + 17i \]
\[ 10 [x(1-3i) - y(1+2i)] = 31 + 17i \]
\[ 10x - 30xi - 10y - 20yi = 31 + 17i \]
Group real and imaginary parts:
\[ (10x - 10y) + i(-30x - 20y) = 31 + 17i \]
Equate parts:
1) \(10x - 10y = 31\)
2) \(-30x - 20y = 17\)
Solve for \(x\) and \(y\):
From (1), \(30x - 30y = 93\) (multiplied by 3).
Add (2) to this result:
\[ (30x - 30y) + (-30x - 20y) = 93 + 17 \]
\[ -50y = 110 \Rightarrow y = -\frac{11}{5} = -2.2 \]
Substitute \(y\) in (1):
\[ 10x - 10(-2.2) = 31 \Rightarrow 10x + 22 = 31 \Rightarrow 10x = 9 \Rightarrow x = 0.9 \]
Calculate the required value:
\[ 10(x - 3y) = 10(0.9 - 3(-2.2)) = 10(0.9 + 6.6) = 10(7.5) = 75 \]
Step 4: Final Answer:
The value of \(10(x-3y)\) is 75.