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

Question

Theodore Amidon · Accepted Answer

#8/65 + 64/65 i#

Two distinct scenarios are possible:

The choice of approach is purely subjective and has little bearing on the outcome.

If it's okay with you, let me try the second strategy.

1) Transform #(5-i)/(2-i)# into #a + bi# form.

To do so, you need to make use of the formula #(x+y)(x-y) = 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#:

#(5-i)/(2-i) = ((5-i)(2+i))/((2-i)(2+i)) = (10 + 3 i - i^2)/(2^2 - i^2)#

... use #i^2 = -1# and simplify...

# = (10 + 3i + 1)/(4 + 1) = (11 + 3i)/5 = 11/5 + 3/5 i#

2) Transform #(3-7i)/(2-3i)# into #a + bi# form.

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

#(3-7i)/(2-3i) = ((3-7i)(2+3i))/((2-3i)(2+3i)) = (6 - 5 i - 21 i^2)/(2^2 - (3i)^2) = (6 - 5i + 21)/(4 + 9)#

#color(white)(xxxxx) = (27-5i)/13 = 27/13 - 5/13 i#

Deduct the pair of terms.

#(5-i)/(2-i) - (3-7i)/(2-3i) = (11/5 + 3/5 i) -(27/13 - 5/13 i)#

#color(white)(xxxxxxxxxxx) = 11/5 - 27/ 13 + (3/5 + 5/13) i#

#color(white)(xxxxxxxxxxx) = 8/65 + 64/65 i#

Anna Allen · Answer

undefined To simplify $\frac{5-i}{2-i} - \frac{3-7i}{2-3i}$ and write it in $a+bi$ form, follow these steps:

1. Find a common denominator for both fractions.
2. Combine the fractions.
3. Write the result in $a+bi$ form by separating the real and imaginary parts.

Solving step by step:

1. Find a common denominator:
   $(2-i)(2+3i)$ is a common denominator for both fractions.

2. Rewrite the fractions with the common denominator:
   $\frac{(5-i)(2+3i)}{(2-i)(2+3i)} - \frac{(3-7i)(2+i)}{(2-i)(2+3i)}$

3. Expand and simplify:
   $\frac{10 + 15i - 2i - 3i^2}{4 + 6i - 2i - 3i^2} - \frac{6 + 2i - 3i - 7i^2}{4 + 6i - 2i - 3i^2}$
   $\frac{10 + 13i - 3(-1)}{4 + 6i - 2i - 3(-1)} - \frac{6 - i - 7(-1)}{4 + 6i - 2i - 3(-1)}$
   $\frac{10 + 13i + 3}{4 + 6i + 2 - 3}$ and $\frac{6 - i + 7}{4 + 6i + 2 - 3}$
   $\frac{13 + 13i}{3 + 6i}$ and $\frac{13 - i}{3 + 6i}$

4. Combine the fractions:
   $\frac{(13 + 13i) - (13 - i)}{3 + 6i}$
   $\frac{13 + 13i - 13 + i}{3 + 6i}$
   $\frac{14i}{3 + 6i}$

5. Multiply by the conjugate of the denominator to rationalize the expression:
   $\frac{14i(3 - 6i)}{(3 + 6i)(3 - 6i)}$
   $\frac{42i - 84i^2}{9 - 18i + 18i - 36i^2}$
   $\frac{42i + 84}{9 + 36}$
   $\frac{42i + 84}{45}$

6. Separate into real and imaginary parts:
   $\frac{84}{45} + \frac{42}{45}i$

7. Simplify the fractions:
   $ \frac{28}{15} + \frac{14}{15}i$

So, $\frac{5-i}{2-i} - \frac{3-7i}{2-3i} = \frac{28}{15} + \frac{14}{15}i$ in $a+bi$ form.