Find the real values of x and y -4+3i-x+yi=8-6i

Mathematics

1 answer:

Marrrta [24]3 years ago

6 0

Answer:

x = - 12, y = - 9

Step-by-step explanation:

Given

- 4 + 3i - x + yi = 8 - 6i

Simplify left side by collecting like terms and equate like coefficients with the terms on the right side, that is

- 4 - x + i(3 + y) = 8 - 6i, thus

3 + y = - 6 ( subtract 3 from both sides ) ← equating imaginary parts

y = - 9, then

- 4 - x = 8 ( add 4 from both sides ) ← equating real parts

- x = 12 ( multiply both sides by - 1 )

x = - 12

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Read 2 more answers

Consider functions f and g.1 + 12f(1) = 12 + 4. – 12for * # 2 and 7 -64.2 – 16. + 1641 +48for a # -12 Which expression is equal

lisabon 2012 [21]

Given the following functions below,

$\begin{gathered} f(x)=\frac{x+12}{x^2+4x-12}\text{ and} \\ g(x)=\frac{4x^2-16x+16}{4x+48} \end{gathered}$

Factorising the denominators of both functions,

Factorising the denominator of f(x),

$\begin{gathered} f(x)=\frac{x+12}{x^2+4x-12}=\frac{x+12}{x^2+6x-2x-12}=\frac{x+12}{x(x+6)-2(x+6)}=\frac{x+12}{(x-2)(x+6)} \\ f(x)=\frac{x+12}{(x-2)(x+6)} \end{gathered}$

Factorising the denominator of g(x),

$\begin{gathered} g(x)=\frac{4x^2-16x+16}{4x+48}=\frac{4(x^2-4x+4)}{4(x+12)} \\ \text{Cancel out 4 from both numerator and denominator} \\ g(x)=\frac{x^2-4x+4}{x+12}=\frac{x^2-2x-2x+4}{x+12}=\frac{x(x-2)-2(x-2)}{x+12}=\frac{(x-2)^2}{x+12} \\ g(x)=\frac{(x-2)^2}{x+12} \end{gathered}$

Multiplying both functions,