Question

3-6i/4-2i simplify the solution
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Answer 1

Answer:

3i/2i

Step-by-step explanation:

3-6i/4-2i

3i(1-2)/2i(2-1)

=3i/2i

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Answer 2

Answer:

Required simple solution is $\frac{6}{5} - \frac{9i}{10}$

Step-by-step explanation:

Given,

$\frac{3 - 6i}{4 - 2i}$

Here we want to simplify it.

$\frac{(3 - 6i)(4 + 2i)}{(4 - 2i)(4 + 2i)} \\ = \frac{12 + 6i - 24i - 12 {i}^{2} }{16 + 8i - 8i - {(2i)}^{2} } \\ = \frac{12 - 18i + 12}{16 - 4 {i}^{2} } \\ = \frac{24 - 18i}{16 + 4} \\ = \frac{24 - 18i}{20} \\ = \frac{24}{20} - \frac{18i}{20} \\ = \frac{6}{5} - \frac{9i}{10}$

Here applied formula,

${i}^{2} = - 1$

This is a problem of Complex number of Algebra.

Some important Algebra formulas:

${(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} \\ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} \\ {(x + y)}^{3} = {x}^{3} + 3 {x}^{2} y + 3x {y}^{2} + {y}^{3} \\ {(x - y)}^{3} = {x}^{3} - 3 {x}^{2} y + 3x {y}^{2} - {y}^{3} \\ {x}^{3} + {y}^{3} = {(x + y)}^{3} - 3xy(x + y) \\ {x}^{3} - {y}^{3} = {(x - y)}^{3} + 3xy(x - y) \\ {x}^{2} - {y}^{2} = (x + y)(x - y) \\ {x}^{2} + {y}^{2} = {(x - y)}^{2} + 2xy \\ {x}^{2} - {y}^{2} = {(x + y)}^{2} - 2xy \\ {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2} ) \\ {x}^{3} + {y}^{3} = (x - + y)( {x}^{2} - xy + {y}^{2} )$

Know more about Algebra,

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