# How do you simplify  (5-i)/(2-i) - (3-7i)/(2-3i) and write in a+bi form?

Dec 3, 2015

$\frac{8}{65} + \frac{64}{65} i$

#### Explanation:

There are two different possibilities:

• either you combine the two fractions into one first and later write it in the $a + b i$ form
• or you first write each of the fractions in $a + b i$ form and later subtract the one from the other

Which approach to pick is a matter of personal preference and doesn't make a big difference.

Let me do the second approach if you don't mind.

1) Transform $\frac{5 - i}{2 - i}$ into $a + b i$ form.

To do so, you need to make use of the formula $\left(x + y\right) \left(x - y\right) = {x}^{2} - {y}^{2}$.
In order to achieve such a situation, you need to multiply both the numerator and the denominator with $2 + i$, the conjugate of $2 - i$:

$\frac{5 - i}{2 - i} = \frac{\left(5 - i\right) \left(2 + i\right)}{\left(2 - i\right) \left(2 + i\right)} = \frac{10 + 3 i - {i}^{2}}{{2}^{2} - {i}^{2}}$

... use ${i}^{2} = - 1$ and simplify...

$= \frac{10 + 3 i + 1}{4 + 1} = \frac{11 + 3 i}{5} = \frac{11}{5} + \frac{3}{5} i$

2) Transform $\frac{3 - 7 i}{2 - 3 i}$ into $a + b i$ form.

The approach is the same: multiply the numerator and the denominator with the conjugate of the denominator, namely $2 + 3 i$, and simplify:

$\frac{3 - 7 i}{2 - 3 i} = \frac{\left(3 - 7 i\right) \left(2 + 3 i\right)}{\left(2 - 3 i\right) \left(2 + 3 i\right)} = \frac{6 - 5 i - 21 {i}^{2}}{{2}^{2} - {\left(3 i\right)}^{2}} = \frac{6 - 5 i + 21}{4 + 9}$

$\textcolor{w h i t e}{\times \times x} = \frac{27 - 5 i}{13} = \frac{27}{13} - \frac{5}{13} i$

3) Subtract the two terms

$\frac{5 - i}{2 - i} - \frac{3 - 7 i}{2 - 3 i} = \left(\frac{11}{5} + \frac{3}{5} i\right) - \left(\frac{27}{13} - \frac{5}{13} i\right)$

$\textcolor{w h i t e}{\times \times \times \times \times x} = \frac{11}{5} - \frac{27}{13} + \left(\frac{3}{5} + \frac{5}{13}\right) i$

$\textcolor{w h i t e}{\times \times \times \times \times x} = \frac{8}{65} + \frac{64}{65} i$