91Ó°ÊÓ

Rewrite the first equation, \( 2x + 2y = 3 \), in slope-intercept form. To do this, solve for \( y \): \[ 2x + 2y = 3 \] Subtract \( 2x \) from both sides: \[ 2y = 3 - 2x \] Divide both sides by 2: \[ y = \frac{3}{2} - x \]Now, graph the equation \( y = \frac{3}{2} - x \) by finding the y-intercept (\( \frac{3}{2} \)) and using the slope (-1) to find another point.

Given the second equation, \( xy = -1 \), solve for one variable in terms of the other. Let's solve for \( y \) in terms of \( x \): \[ y = \frac{-1}{x} \]

Graph the equation \( y = \frac{-1}{x} \). This is a hyperbola with asymptotes along the x and y axes. Plot a few points to sketch its shape. For example: If \( x = 1 \), then \( y = -1 \).If \( x = -1 \), then \( y = 1 \).If \( x = 2 \), then \( y = \frac{-1}{2} \).If \( x = -2 \), then \( y = \frac{1}{2} \).

Determine the intersection points of the two graphs from the previous steps. To find the intersection points algebraically, set \( \frac{3}{2} - x = \frac{-1}{x} \) and solve for \( x \):\[ \frac{3}{2} - x = \frac{-1}{x} \]Multiply both sides by \( x \):\[ x \frac{3}{2} - x^2 = -1 \]Rearrange to form a quadratic equation:\[ x^2 - \frac{3}{2}x - 1 = 0 \]Solve this quadratic equation using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -\frac{3}{2} \), and \( c = -1 \):\[ x = \frac{\frac{3}{2} \pm \sqrt{(\frac{3}{2})^2 + 4}}{2} \]\[ = \frac{\frac{3}{2} \pm \sqrt{\frac{9}{4} + 4}}{2} \]\[ = \frac{\frac{3}{2} \pm \sqrt{\frac{9}{4} + \frac{16}{4}}}{2} \]\[ = \frac{\frac{3}{2} \pm \sqrt{\frac{25}{4}}}{2} \]\[ = \frac{\frac{3}{2} \pm \frac{5}{2}}{2} \]This yields two values for \( x \):\[ x = \frac{4}{2} = 2 \] or \[ x = \frac{-1}{2} \].Using these \( x \) values in \( y = \frac{3}{2} - x \), we find the corresponding \( y \) values: For \( x = 2 \):\[ y = \frac{3}{2} - 2 = -\frac{1}{2} \]For \( x = -\frac{1}{2} \):\[ y = \frac{3}{2} - (-\frac{1}{2}) = 2 \].So, the intersection points are \( (2, -\frac{1}{2}) \) and \( (-\frac{1}{2}, 2) \).