$(2+3i)x² - (3-2i)y = 2x - 3y + 5i$
Where "i" is $\sqrt-1$
2026-04-03 07:33:37.1775201617
Solve for x & y for this complex equation.
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Two complex numbers are equal if and only if the real parts are equal and the imaginary parts are also equal.
In your problem we have $$ 2x^2 -3y = 2x-3y$$ and $$ 3x^2+2y =5$$.
solve for $x$ and $y$ to get $$ x=0, y=5/2$$ or $$x=1, y=1$$