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[x2/y2]-3[-x/2y]=[-2/3] find x,y |
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Answer» \(\begin{bmatrix}x^2 \\[0.3em]y^2\end{bmatrix}\) - 3\(\begin{bmatrix}- x \\[0.3em]2y\end{bmatrix}\)= \(\begin{bmatrix}- 2 \\[0.3em]3\end{bmatrix}\) ⇒ \(\begin{bmatrix}x^2 + 3x\\[0.3em]y^2 -6y\end{bmatrix}\) = \(\begin{bmatrix}- 2 \\[0.3em]3\end{bmatrix}\) ⇒ x2 + 3x = -2 & y2 - 6y = 3 ⇒ x2 + 3x + 2 = 0 & y2 - 6y - 3 = 0 ⇒ (x + 1) (x + 2) = 0 & y = \(\frac{6±\sqrt{36+12}}{2}\) = \(\frac{6±4\sqrt 3}{2}\) = 3 ± √3 ⇒ x = -1 or x = -2 & y = 3 + √3 or y = 3 - √3 |
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