Find the values of the real numbers x and y

Question

x/(2−i) + yi/(i+3) = 2/(1+i) here "i" = iota how can we find real part? when iota imaginary is also in equation! kindly solve.

Answer 1

Remembering that when you add or subtract fractions the denominators must be the same. In this case the common denominator would be (2-i)(3+i)(1+i). Once we have all the fractions with the same denominators, all we need to look at is the numerators. So, whatever you have to multiply the denominator of a fraction you, also, have to multiply the numerator.

Therefore, we get:

x(3+i)(1+i)+yi(2-i)(1+i)=2(2-i)(3+i)

x(3+4i+i²)+yi(2+i-i²)=2(6-i-i²)

x(3+4i-1)+yi(2+i+1)=2(6-i+1)

x(2+4i)+yi(3+i)=2(7-i)

2x+4xi+3yi+yi²=14-2i

2x+4xi+3yi-y=14-2i

(2x-y)+(4x+3y)i=14-2i

Can you see now that both sides are in the form of a+bi? 

This gives us:  2x-y=14

                      4x+3y=-2

Using substitution, solve 2x-y=14 to get y=2x-14. Putting this value in for y in the 2nd equation and you get:

4x+3(2x-14)=-2

4x+6x-42=-2       combine like terms

10x=-2+42          add 42 to both sides

10x=40               simplify

x=4                     solve for x

Put this value for x in either one of the original equations to solve for y:

2x-y=14

2(4)-y=14

8-y=14

8-14=y

y=-6

Your answers are x=4, y=-6. Putting these values in the original equation and solving will be the check to see if your answers are correct.

Answer 2

x / (2-i)[(2+i)/(2+i)] + yi/(i+3)[(-i+3)/(-i+3)] = 2/(1+i)[(1-i)/(1-i)]

 

(2x+xi)/5 + (y+3yi)/10 = (2-2i)/2

 

Multiply both sides by the LCD, 10:

 

    2(2x+xi) + (y+3yi) = 5(2-2i)

 

    (4x+y) + (2x+3y)i = 10 - 10i

 

So, 4x+y = 10 and 2x+3y = -10

 

Solving for x and y, we get y = -6 and x = 4