# What is a if (2-i)(a-bi) = 2+9i ?

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We have to find a given that (2-i)(a-bi) = 2+9i.

Now, (2-i)(a-bi) = 2+9i

=> 2a -ia -2bi + bi^2 = 2+ 9i

=> 2a - ia - 2bi - b = 2+ 9i

=> 2a -b -i ( a+ 2b) = 2+ 9i

Equate the real and imaginary coefficients.

We get 2a -b = 2 and a + 2b = -9

Now as 2a -b = 2 => b = 2a - 2

substitute this in a + 2b = -9

=> a + 2*( 2a - 2) = -9

=> a + 4a - 4 = -9

=> 5a = -5

=> a = -5/5 = -1

**Therefore the required value of a is -1.**

We'll remove the brackets from the left side

(2-i)(a-bi) = 2+9i

2a - 2bi - ai + bi^2 = 2 + 9i

But i^2 = -1

We'll re-write with respect to this:

2a - b - 2bi - ai = 2 + 9i

We'll combine real parts and imaginary parts:

(2a - b) + i(-2b - a) = 2 + 9i

The real and imaginary parts from both sides have to be equal:

2a - b = 2 (1)

-a - 2b = 9 (2)

We'll multiply (2) by 2:

-2a - 4b = 18

(1)+(2):

2a - b -2a - 4b = 2 + 18

We'll combine and eliminate like terms:

-5b = 20

**b = -4**

2a + 4 = 2

2a = 2 - 4

2a = -2

**a = -1**