Question

Solve the following equation for x,y£R
(x+iy) (5+6i)=2+3i
​

Answer 1

Answer:

x = \dfrac{28}{61}

61

28

and y = \dfrac{3}{61}

61

3

Step-by-step explanation:

Since we have given that

(x +iy)(5 +6i) = 2 +3i

We need to find the value of x and y.

So, it becomes,

\begin{gathered}x+iy=\dfrac{2+3i}{5+6i}\\\\x+iy=\dfrac{2+3i(5-6i)}{(5+6i)(5-6i)}\\\\x+iy=\dfrac{10-12i+15i+18}{25+36}\\\\x+iy=\dfrac{28+3i}{61}\\\\x+iy=\dfrac{28}{61}+\dfrac{3}{61}i\end{gathered}

x+iy=

5+6i

2+3i

x+iy=

(5+6i)(5−6i)

2+3i(5−6i)

x+iy=

25+36

10−12i+15i+18

x+iy=

61

28+3i

x+iy=

61

28

+ x = \dfrac{28}{61}

61

28

and y = \dfrac{3}{61}

61

3

Step-by-step explanation:

Since we have given that

(x +iy)(5 +6i) = 2 +3i

We need to find the value of x and y.

So, it becomes,

\begin{gathered}x+iy=\dfrac{2+3i}{5+6i}\\\\x+iy=\dfrac{2+3i(5-6i)}{(5+6i)(5-6i)}\\\\x+iy=\dfrac{10-12i+15i+18}{25+36}\\\\x+iy=\dfrac{28+3i}{61}\\\\x+iy=\dfrac{28}{61}+\dfrac{3}{61}i\end{gathered}

x+iy=

5+6i

2+3i

x+iy=

(5+6i)(5−6i)

2+3i(5−6i)

x+iy=

25+36

10−12i+15i+18

x+iy=

61

28+3i

x+iy=

61

28

+

61

3

i

hence, x = \dfrac{28}{61}

61

28

and y = \dfrac{3}{61}

61

3

# learn more:

(x+iy)(2-3i)=4+i

61

3

i

hence, x = \dfrac{28}{61}

61

28

and y = \dfrac{3}{61}

61

3

# learn more:

(x+iy)(2-3i)=4+i

Answer 2

Answer:

By multiplying LHS

5x+6xi+5iy-6y since i^2=-1

5x-6y+i(6x+5y)=2+3i

by comparing 5x-6y=2 and 6x+5y=3

by solving these two linear equations we get x and y