Question

The real values of x and y for which the equation is $\left( x+iy \right)\left( 2-3i \right)=4+i$ is satisfied are:
a). $x=\dfrac{5}{13},y=\dfrac{8}{13}$
b). $x=\dfrac{8}{13},y=\dfrac{5}{13}$
c). $x=\dfrac{5}{13},y=\dfrac{14}{13}$
d). None of these

Answer 1

Hint: We will first multiply both the terms on the LHS of the equation. This would give us the real part and the imaginary part as a linear term of two variables x and y. Comparing the real and imaginary parts of LHS with that of RHS, This will be two simultaneous linear equations. Solving these two equations would give us x and y values.

Complete step-by-step solution -
Let us consider the LHS.
$=\left( x+iy \right)\left( 2-3i \right)$
Multiplying the terms we get,
LHS $=\left( 2\times x \right)+\left( -3i\times x \right)+\left( 2\times iy \right)+\left( -3i\times iy \right)$ .
LHS $=2x-3xi+2yi-3{{i}^{2}}y$ .
Since ${{i}^{2}}=-1$ , we get,
LHS $=2x-3xi+2yi-3\left( -1 \right)y$ .
LHS$=2x-3xi+2yi+3y$
Now grouping all the real parts in one term and all the imaginary parts in the other, we get,
LHS $=\left( 2x+3y \right)+\left( -3x+2y \right)i$
Now, consider RHS.
RHS$=4+i$ .
Therefore, on comparing the LHS and RHS we get,
$2x+3y=4$ and $-3x+2y=1$
Let us solve these two simultaneous equation
$\begin{align}
  & 2x+3y=4.................\times 3 \\
 & -3x+2y=1...............\times 2 \\
\end{align}$
We now get,
$\begin{align}
  & 6x+9y=12 \\
 & -6x+4y=2 \\
\end{align}$
Adding both the equations, we get,
$13y=14$
Cross multiplying,
$y=\dfrac{14}{13}$ .
Substituting this y value in $2x+3y=4$ we get,
$\begin{align}
  & 2x+3\left( \dfrac{14}{13} \right)=4 \\
 & 2x+\dfrac{42}{13}=4 \\
 & 2x=4-\dfrac{42}{13} \\
 & 2x=\dfrac{52-42}{13} \\
 & 2x=\dfrac{10}{13} \\
 & x=\dfrac{5}{13}. \\
\end{align}$
Thus, $x=\dfrac{5}{13}$ and $y=\dfrac{14}{13}$
Thus, option (c) is correct.

Note: There is an alternate method to solve this problem.
Consider the equation.
$\left( x+iy \right)\left( 2-3i \right)=4+i$ .
Cross multiplying $2-3i$ , we get,
$x+iy=\dfrac{4+i}{2-3i}$ .
Multiplying and dividing by the conjugate of $2-3i$ that is $2+3i$ we get,
$x+iy=\dfrac{4+i}{2-3i}\times \dfrac{2+3i}{2+3i}$ .
Multiplying further,
$\begin{align}
  & x+iy=\dfrac{\left( 4\times 2 \right)+\left( 4\times 3i \right)+\left( 2\times i \right)+\left( 3i\times i \right)}{\left( 2\times 2 \right)+\left( 2\times 3i \right)+\left( -3i\times 2 \right)+\left( -3i\times 3i \right)} \\
 & x+iy=\dfrac{8+12i+2i+3{{i}^{2}}}{4+6i-6i-9{{i}^{2}}} \\
\end{align}$
Now we know ${{i}^{2}}=-1$ .
Thus, $\begin{align}
  & x+iy=\dfrac{\left( 8-3 \right)+14i}{4+9} \\
 & x+iy=\dfrac{5+14i}{13} \\
 & x+iy=\dfrac{5}{13}+\dfrac{14i}{13} \\
\end{align}$
Comparing, we get to know, $x=\dfrac{5}{13}$ and $y=\dfrac{14}{13}$ .