Question

Solve the following equations for real x and y.
\[3+5i+x-yi=6-2i\]

Answer 1

Hint: First of all, transpose all the terms of the given equation to LHS and separate the real and imaginary parts of the equation. Now equate the real part and imaginary part of the equation to zero individually to get the values of x and y.

Complete step-by-step answer:
In this question, we have to solve the equation \[3+5i+x-yi=6-2i\] for real values of x and y. First of all, let us consider the equation given in the question.
\[3+5i+x-yi=6-2i\]
Let us transpose all the terms to the left-hand side (LHS) of the above equation, we get,
\[3+5i+x-yi-6+2i=0\]
By separating the real and imaginary terms of the above equation, we get,
\[\left( 3+x-6 \right)+\left( 5i-yi+2i \right)=0\]
\[\Rightarrow \left( 3+x-6 \right)+\left( 5-y+2 \right)i=0\]
By simplifying the above equation, we get,
\[\Rightarrow \left( -3+x \right)+\left( 7-y \right)i=0\]
We know that in any equation the left-hand side is equal to the right-hand side. We can see that in the RHS of the above equation, both the real part and imaginary parts are zero. So, we can write the above equation as,
\[\Rightarrow \left( -3+x \right)+\left( 7-y \right)i=\left( 0 \right)+\left( 0 \right)i.....\left( i \right)\]
Now, by equation real part of the LHS to the real part of the RHS of the above equation, we get, – 3 + x = 0
By adding 3 on both the sides of the above equation, we get x = 3.
Now by equating the imaginary part of the LHS to the imaginary part of the RHS of the equation (i), we get, (7 – y) = 0.
By subtracting 7 from both the sides of the above equation, we get,
\[-y=-7\]
So, we get, y = 7.
Hence, we get the values of x and y as 3 and 7 respectively.

Note: Here, students must note that the values of x and y are real. Also, if we have an equation of the form A + iB = 0 then we get A = 0 and B = 0. In this question, students can cross-check their answers by substituting the values of x and y in the given equation as follows:
Our equation is \[3+5i+x-yi=6-2i\]. By substituting the values of x = 3 and y = 7 in the above equation, we get,
\[3+5i+3-7i=6-2i\]
\[6-2i=6-2i\]
LHS = RHS
Here, we get LHS = RHS. Hence, our answer is correct.