Let x and y be real numbers such that 50((2x)/(1+3i)-(y)/(1-2i))=31+17i,i=-1.

Question

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

Accepted Answer

$50(\frac{2x}{1+3i}-\frac{y}{1-2i})=31+17i$

We need to find $10(x-3y)$.

First, simplify the fractions by rationalising the denominators.

$\frac{2x}{1+3i}=\frac{2x(1-3i)}{(1+3i)(1-3i)}$

$=\frac{2x(1-3i)}{1+9}=\frac{2x(1-3i)}{10}$

$=\frac{x}{5}(1-3i)$

Similarly,

$\frac{y}{1-2i}=\frac{y(1+2i)}{(1-2i)(1+2i)}$

$=\frac{y(1+2i)}{1+4}=\frac{y(1+2i)}{5}$

Now substitute into the given equation:

$50(\frac{x}{5}(1-3i)-\frac{y}{5}(1+2i))=31+17i$

Take $\frac{1}{5}$ common inside:

$50\cdot \frac{1}{5}[x(1-3i)-y(1+2i)]=31+17i$

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

Expand:

$10[x-3xi-y-2yi]=31+17i$

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

So,

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

Now compare real and imaginary parts:

$10(x-y)=31$

$10(-3x-2y)=17$

Thus,

$x-y=\frac{31}{10}$

$-3x-2y=\frac{17}{10}$

Now solve these equations.

From the first equation,

$x=y+\frac{31}{10}$

Substitute into the second:

$-3(y+\frac{31}{10})-2y=\frac{17}{10}$

$-3y-\frac{93}{10}-2y=\frac{17}{10}$

$-5y=\frac{17}{10}+\frac{93}{10}$

$-5y=\frac{110}{10}=11$

$y=-\frac{11}{5}$

Then

$x=y+\frac{31}{10}=-\frac{11}{5}+\frac{31}{10}$

$x=-\frac{22}{10}+\frac{31}{10}=\frac{9}{10}$

Now find:

$10(x-3y)=10(\frac{9}{10}-3(-\frac{11}{5}))$

$=10(\frac{9}{10}+\frac{33}{5})$

$=10(\frac{9}{10}+\frac{66}{10})$

$=10\cdot \frac{75}{10}=75$

Therefore, the value of $10(x-3y)$ is

$75$

So, the correct option is .

Solutions · 1