Solving Complex Number Equation
We are given the equation: $50\left(\frac{2x}{1+3i} - \frac{y}{1-2i}\right) = 31 + 17i$ We need to find the value of $10(x-3y)$. $x$ and $y$ are real numbers.
Step 1: Simplify Complex Fractions
First, rationalize the denominators:
- $\frac{2x}{1+3i} = \frac{2x(1-3i)}{(1+3i)(1-3i)} = \frac{2x(1-3i)}{1^2 + 3^2} = \frac{2x(1-3i)}{10} = \frac{x(1-3i)}{5}$
- $\frac{y}{1-2i} = \frac{y(1+2i)}{(1-2i)(1+2i)} = \frac{y(1+2i)}{1^2 + 2^2} = \frac{y(1+2i)}{5}$
Step 2: Substitute and Simplify the Equation
Substitute the simplified fractions back into the main equation:
$50\left(\frac{x(1-3i)}{5} - \frac{y(1+2i)}{5}\right) = 31 + 17i$ $10\left( x(1-3i) - y(1+2i) \right) = 31 + 17i$ $10\left( x - 3xi - y - 2yi \right) = 31 + 17i$Group the real and imaginary terms:
$10\left( (x-y) - (3x+2y)i \right) = 31 + 17i$ $(10x - 10y) - (30x + 20y)i = 31 + 17i$Step 3: Equate Real and Imaginary Parts
Equate the real parts and the imaginary parts of the equation:
- Real part: $10x - 10y = 31$ (Equation 1)
- Imaginary part: $-(30x + 20y) = 17 \implies 30x + 20y = -17$ (Equation 2)
Step 4: Solve the System of Linear Equations
Solve the system formed by Equation 1 and Equation 2:
- Multiply Equation 1 by 3: $30x - 30y = 93$ (Equation 3)
- Subtract Equation 2 from Equation 3: $(30x - 30y) - (30x + 20y) = 93 - (-17)$ $-50y = 110$ $y = -\frac{110}{50} = -\frac{11}{5}$
- Substitute $y = -\frac{11}{5}$ into Equation 1: $10x - 10\left(-\frac{11}{5}\right) = 31$ $10x + 22 = 31$ $10x = 9$ $x = \frac{9}{10}$
Step 5: Calculate the Target Expression
Calculate the value of $10(x-3y)$ using the found values of $x$ and $y$:
$10(x-3y) = 10\left(\frac{9}{10} - 3\left(-\frac{11}{5}\right)\right)$ $10\left(\frac{9}{10} + \frac{33}{5}\right)$ $10\left(\frac{9}{10} + \frac{66}{10}\right)$ $10\left(\frac{9 + 66}{10}\right)$ $10\left(\frac{75}{10}\right)$ $10(x-3y) = 75$Similar Questions
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