Problems

Show that \[ x^2-y^2 +x+3y-2 = (x-y+2)(x+y-1) \] and hence, or otherwise, indicate by means of a sketch the region of the \(x\)-\(y\) plane for which $$ x^2-y^2 +x+3y>2. $$ Sketch also the region of the \(x\)-\(y\) plane for which $$ x^2-4y^2 +3x-2y<-2. $$ Give the coordinates of a point for which both inequalities are satisfied or explain why no such point exists.

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Solution: \begin{align*} && (x-y+2)(x+y-1) &= (x-y)(x+y)-(x-y)+2(x+y)-2 \\ &&&= x^2-y^2+x+3y-2 \end{align*}

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\begin{align*} x^2-4y^2 +3x-2y+2 &= (x - 2 y + 1) (x + 2 y + 2) \end{align*}

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Consider the point \(x = 0, y = \frac32\), then \(\frac92 - \frac94 = \frac94 > 2\) and \(-4\cdot\frac94-2\cdot \frac32 = -12 < -2\) so this is an example of a point in both regions