To find the values of x and y in the equation \(3x + i(4x - y) = 6 - i\), we can equate the real and imaginary parts on both sides of the equation.
Equating the real parts:
3x = 6
Dividing both sides by 3, we get:
x = 2
Equating the imaginary parts:
\(i(4x - y) = -i\)
Multiplying both sides by -i, we get:
\(4x - y = -1\)
Substituting the value of x from the first equation, we have:
\(4(2) - y = -1\)
\(8 - y = -1\)
Subtracting 8 from both sides, we get:
\(-y = -9\)
Dividing both sides by -1, we get:
y = 9
Therefore, the values of x and y in the equation \(3x + i(4x - y) = 6 - i\) are x = 2 and y = 9 (option B).
The portion of the line \( 4x + 5y = 20 \) in the first quadrant is trisected by the lines \( L_1 \) and \( L_2 \) passing through the origin. The tangent of an angle between the lines \( L_1 \) and \( L_2 \) is:
If f(x) = ex, h(x) = (fof) (x), then \(\frac{h'(x)}{h'(x)}\) =