Answer to Question #251906 in Algebra for moe

Question

Consider the equation $\frac{x}{2+i}+\frac{iy}{2-i}=\frac{3}{1-2i}$ , where x and y are real numbers. Which of the following are correct?

1. x=−4 and y =-5
2. x=5 and y=4
3. x and y satisfy the pair of equations  2x−y=3 and -x+2y=6
4. x=4 and y=5
5. x and y satisfy the pair of equations 2x+y=3 and x-2y=-6

Answer 1

1)

L.H.S=$\frac{x}{2+i}+\frac{iy}{2-i}$

put x=-4 and y=-5.

$\frac{-4}{2+i}+\frac{-i5}{2-i}\\ \frac{-4(2-i)-5i(2+i)}{(2+i)(2-i)}\\ \frac{-4(2-i)-5i(2+i)}{2^2-i^2}\\ \frac{-3-6i}{5}\\$

Not equal to RHS.

2)

put x=5 and y=4.

$LHS=\frac{5}{2+i}+\frac{i4}{2-i}\\ \frac{5(2-i)+4i(2+i)}{(2+i)(2-i)}\\ \frac{5(2-i)+4i(2+i)}{2^2-i^2}\\ \frac{6+3i}{5}\\$

Not equal to RHS.

3)

solving equation

2x-y=3 and -x+2y=-6, we get x=4, y=5.

$LhS=\frac{4}{2+i}+\frac{i5}{2-i}\\ \frac{4(2-i)+5i(2+i)}{(2+i)(2-i)}\\ \frac{4(2-i)+5i(2+i)}{2^2-i^2}\\ \frac{3+6i}{5}\\$

Not equal to RHS.

4)

putx=4,y=5.

$LhS=\frac{4}{2+i}+\frac{i5}{2-i}\\ \frac{4(2-i)+5i(2+i)}{(2+i)(2-i)}\\ \frac{4(2-i)+5i(2+i)}{2^2-i^2}\\ \frac{3+6i}{5}\\$

Not equal to RHS.

5)

Solving equations 2x+y=3 and x-2y=-6, we get x=0,y=3.

$LhS=\frac{0}{2+i}+\frac{3i}{2-i}\\ \frac{3i}{2-i}\\ \text{By rationalise, we get}\\ \frac{-3+6i}{5}\\$

Not equal to RHS.

No options are correct.