Question

((4-2i)(2+l)+(3+i)·x=4-2i​

Answer 1

(4-2i)(2+i)+(3+i)·x=4-2i

8+4i-4i-2i^2+3x+xi+2i=4

-2i^2+2i+xi+3x=-4

-2i(i-1)+x (i+3)=-4

Answer 2

Let's solve the equation step by step.

Given equation: $(4 - 2i)(2 + 1) + (3 + i) \cdot x = 4 - 2i$

Step 1: Simplify the expressions on the left-hand side (LHS) of the equation: $(4 - 2i)(2 + 1) = (4 - 2i)(3) = 12 - 6i$

Step 2: Rewrite the equation with the simplified expressions: $12 - 6i + (3 + i) \cdot x = 4 - 2i$

Step 3: Move the constant term to the right-hand side (RHS) of the equation: $(3 + i) \cdot x = 4 - 2i - (12 - 6i)$

Step 4: Simplify the terms on the RHS: $(3 + i) \cdot x = 4 - 2i - 12 + 6i$

Step 5: Continue simplifying: $(3 + i) \cdot x = -8 + 4i$

Step 6: Now, isolate the variable x by dividing both sides by the coefficient of x, which is (3 + i): $x = \frac{-8 + 4i}{3 + i}$

Step 7: Rationalize the denominator by multiplying the numerator and denominator by the conjugate of (3 + i), which is (3 - i): $x = \frac{(-8 + 4i)(3 - i)}{(3 + i)(3 - i)}$

Step 8: Multiply the numerator and denominator: $x = \frac{-24 + 8i + 12i - 4i^2}{9 - i^2}$

Step 9: Simplify further (remembering that $i^2 = -1$): $x = \frac{-24 + 20i - 4(-1)}{9 - (-1)}$ $x = \frac{-24 + 20i + 4}{9 + 1}$ $x = \frac{-20 + 20i}{10}$

Step 10: Factor out 20 from the numerator: $x = \frac{20(-1 + i)}{10}$

Step 11: Simplify the fraction by canceling out the common factor of 10: $x = 2(-1 + i)$

So, the solution to the equation is $x = 2(-1 + i)$.