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Sati [7]
3 years ago
10

2x-3yi=14+12i how do I find x and y

Mathematics
2 answers:
vladimir1956 [14]3 years ago
7 0
2x-3yi=14+12i 
=> 2x=14; -3yi=12i => x=7; y=-4
egoroff_w [7]3 years ago
4 0

Answer:

The answers are:

x=7\\y=-4

Step-by-step explanation:

In order to determine the answer, we have to know about complex numbers.

The complex numbers have one difference respect on real numbers, they have an imaginary unit i.

In general, a complex number has the next form:

a+b*i

"a" and "b" are real numbers.

We can not add the imaginary part (where it is the imaginary unit) with the real part.

In this case, we have an equality, where in the right side, we can see a complex number and in the left side, we can see two variables and the imaginary unit. We just have to build two equation: one for the real part and other one for the imaginary part.

2x=14\\x=7\\\\-3y=12\\y=-4

Finally, the values of the variables to fulfil the equality are:

x=7\\y=-4

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Answer:

x = √17 and x = -√17

Step-by-step explanation:

We have the equation:

\frac{3}{x + 4}  - \frac{1}{x + 3}  = \frac{x + 9}{(x^2 + 7x + 12)}

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Then we can first multiply both sides by (x + 4) to get:

\frac{3*(x + 4)}{x + 4}  - \frac{(x + 4)}{x + 3}  = \frac{(x + 9)*(x + 4)}{(x^2 + 7x + 12)}

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Now we can multiply both sides by (x + 3)

3*(x + 3)  - \frac{(x + 4)*(x+3)}{x + 3}  = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}

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(2*x + 5)  = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}

Now we can multiply both sides by (x^2 + 7*x + 12)

(2*x + 5)*(x^2 + 7x + 12)  = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}*(x^2 + 7x + 12)

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This means that we can take a factor (x + 3) out, so we can rewrite our equation as:

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Then the other two solutions are given by:

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And neither of these have problems in the denominators, so we can conclude that the solutions are:

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