Q. 21 B5.0( 2 Votes )

# Find the real values of x and y for which:(x + iy) (3 – 2i) = (12 + 5i)

x(3 – 2i) + iy(3 – 2i) = 12 + 5i

3x – 2ix + 3iy – 2i2y = 12 + 5i

3x + i(-2x + 3y) – 2(-1)y = 12 + 5i [ i2 = -1]

3x + i(-2x + 3y) + 2y = 12 + 5i

(3x + 2y) + i(-2x + 3y) = 12 + 5i

Comparing the real parts, we get

3x + 2y = 12 …(i)

Comparing the imaginary parts, we get

–2x + 3y = 5 …(ii)

Solving eq. (i) and (ii) to find the value of x and y

Multiply eq. (i) by 2 and eq. (ii) by 3, we get

6x + 4y = 24 …(iii)

–6x + 9y = 15 …(iv)

Adding eq. (iii) and (iv), we get

6x + 4y – 6x + 9y = 24 + 15

13y = 39

y = 3

Putting the value of y = 3 in eq. (i), we get

3x + 2(3) = 12

3x + 6 = 12

3x = 12 – 6

3x = 6

x = 2

Hence, the value of x = 2 and y = 3

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Evaluate:

(i) (iii) .

RS Aggarwal - Mathematics