Question

Find the real values of x and y if
\[\left( 3x-2iy \right){{\left( 2+i \right)}^{2}}=10\left( 1+i \right)\]

Answer 1

Hint: First of all simplify the equation by using \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]. Now transpose all the terms to the LHS and separate real and imaginary parts of the equation. Now equate real and imaginary parts to 0 individually to get the values of x and y.


Complete step-by-step solution -

In this question, we have to solve the equation \[\left( 3x-2iy \right){{\left( 2+i \right)}^{2}}=10\left( 1+i \right)\] for real values of x and y. Let us consider the equation given in the question.
\[\left( 3x-2iy \right){{\left( 2+i \right)}^{2}}=10\left( 1+i \right)\]
We know that \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]. By using this in the above equation, we get,
\[\left( 3x-2iy \right)\left( {{\left( 2 \right)}^{2}}+{{\left( i \right)}^{2}}+4i \right)=10\left( 1+i \right)\]
We know that \[i=\sqrt{-1}\]. So, we get, \[{{i}^{2}}=-1\]. So, by substituting \[{{i}^{2}}=-1\] in the above equation, we get,
\[\Rightarrow \left( 3x-2iy \right)\left( 4+\left( -1 \right)+4i \right)=10+10i\]
\[\Rightarrow \left( 3x-2iy \right)\left( 3+4i \right)=10+10i\]
By simplifying the above equation, we get,
\[9x+12xi-6iy-8{{\left( i \right)}^{2}}y=10+10i\]
By substituting \[{{i}^{2}}=-1\] in the above equation, we get,
\[\Rightarrow 9x+12xi-6iy-8\left( -1 \right)y=10+10i\]
\[\Rightarrow 9x+12xi-6iy+8y=10+10i\]
By transposing all the terms of the above equation to LHS, we get,
\[\Rightarrow 9x+12xi-6iy+8y-10-10i=0\]
Now, by separating real and imaginary parts of the above equation, we get,
\[\left( 9x+8y-10 \right)+\left( 12x-6y-10 \right)i=0\]
We know that in any equation, LHS = RHS. Also, we can see that in the RHS of the above equation, real and imaginary parts are zero. So, we can write the above equation as
\[\left( 9x+8y-10 \right)+\left( 12x-6y-10 \right)i=0+0i.....\left( i \right)\]
Now, by equating real part of the LHS to real part of the RHS of the above equation, we get,
\[9x+8y-10=0....\left( ii \right)\]
Also, by equating the imaginary part of the LHS to the imaginary part of the RHS of equation (i), we get,
\[12x-6y-10=0\]
\[\Rightarrow 6x-3y-5=0....\left( iii \right)\]
By solving the above equation, we get,
\[-3y=-6x+5\]
\[\Rightarrow 3y=6x-5\]
\[y=\dfrac{6x-5}{3}\]
Now, by substituting \[y=\dfrac{6x-5}{3}\] in equation (ii), we get,
\[9x+\dfrac{8}{3}\left( 6x-5 \right)-10=0\]
\[9x+\dfrac{48}{3}x-\dfrac{40}{3}-\dfrac{10}{1}=0\]
\[9x+16x-\dfrac{\left( 40+30 \right)}{3}=0\]
\[25x=\dfrac{70}{3}\]
\[x=\dfrac{70}{3\times 25}=\dfrac{14}{15}\]
So, we get, \[x=\dfrac{14}{15}\].
By substituting the value of \[x=\dfrac{14}{15}\] in equation (v), we get,
\[y=\dfrac{6\left( \dfrac{14}{15} \right)-5}{3}\]
\[\Rightarrow y=\dfrac{\dfrac{\left( 2.14 \right)}{5}-5}{3}\]
\[\Rightarrow y=\dfrac{\dfrac{28}{5}-5}{3}\]
\[\Rightarrow y=\dfrac{\dfrac{28-25}{5}}{3}\]
\[y=\dfrac{3}{5.3}=\dfrac{1}{5}\]
So, we get the values of x and y as \[\dfrac{14}{15}\] and \[\dfrac{1}{5}\] respectively.

Note: Here students should note that the values of x and y must be real. Also, it is very important to first properly simplify the equation and recheck every step to avoid any type of mistake. Students can also cross check their answer by substituting the values of x and y in the given equation as follows:
\[\left( 3x-2iy \right){{\left( 2+i \right)}^{2}}=10\left( 1+i \right)\]
By substituting \[x=\dfrac{14}{15}\] and \[y=\dfrac{1}{5}\], we get,
\[\left[ 3\left( \dfrac{14}{15} \right)-2i\left( \dfrac{1}{5} \right) \right]{{\left( 2+i \right)}^{2}}=10+10i\]
\[\left( \dfrac{14}{5}-\dfrac{2i}{5} \right)\left( 4+{{i}^{2}}+4i \right)=10+10i\]
\[\dfrac{\left( 14-2i \right)}{5}\left( 4-1+4i \right)=10+10i\]
\[\dfrac{\left( 14-2i \right)\left( 3+4i \right)}{5}=10+10i\]
\[\dfrac{42+56i-6i-8{{i}^{2}}}{5}=10+10i\]
\[\dfrac{50+50i}{5}=10+10i\]
\[10+10i=10+10i\]
Hence, LHS = RHS. Therefore, our answer is correct.