Graph each inequality. \( \begin{array}{lll}\text { 1. } yx+12 & \text { 3. } 2 x+2 y<18 \\ \text { 4. } y-2>-2(x-1) & \text { 5. } 5 x+y>10 & \text { 6. }-24 x+8 y \geq-48 \\ \text { 7. } 2 y+6 \geq 0 & \text { 8. } \frac{1}{2} x+1<3 & \text { 9. } \frac{2}{3} x-\frac{10}{3}>-4\end{array} \)

1. For the inequality \[ y < x - 3, \] write the boundary as \[ y = x - 3. \] Because the inequality is strict (\(<\)), the line is drawn dashed. The region satisfying the inequality is the area below this line. --- 2. For the inequality \[ y > x + 12, \] write the boundary as \[ y = x + 12. \] Since the inequality is strict (\(>\)), use a dashed line, and shade the region above the line. --- 3. For the inequality \[ 2x + 2y < 18, \] divide both sides by 2 to obtain \[ x + y < 9. \] The boundary line is \[ x + y = 9, \] or written in slope–intercept form, \[ y = 9 - x. \] Draw this line dashed (because of the \(<\) sign) and shade the region below it (points where \(x+y\) is less than 9). --- 4. For the inequality \[ y - 2 > -2(x - 1), \] first expand the right–side: \[ y - 2 > -2x + 2. \] Then add 2 to both sides: \[ y > -2x + 4. \] The boundary line is \[ y = -2x + 4, \] drawn dashed (since the inequality is strict). Shade the region above the line (where \(y\) is greater than \(-2x+4\)). --- 5. For the inequality \[ 5x + y > 10, \] solve for \(y\) by subtracting \(5x\): \[ y > -5x + 10. \] The boundary line is \[ y = -5x + 10. \] Since the inequality is strict (\(>\)), draw the line dashed and shade the region above it. --- 6. For the inequality \[ -24x + 8y \geq -48, \] divide both sides by 8 (note that division by a positive number does not change the inequality): \[ -3x + y \geq -6. \] Solve for \(y\) by adding \(3x\) to both sides: \[ y \geq 3x - 6. \] The boundary line is \[ y = 3x - 6. \] Because the inequality is non–strict (\(\geq\)), draw the line solid and shade the region above the line. --- 7. For the inequality \[ 2y + 6 \geq 0, \] solve for \(y\) by subtracting 6: \[ 2y \geq -6, \] and then dividing by 2: \[ y \geq -3. \] The boundary is the horizontal line \[ y = -3. \] Since the inequality is \(\geq\), draw it as a solid line and shade the region above the line. --- 8. For the inequality \[ \frac{1}{2}x + 1 < 3, \] solve for \(x\). Subtract 1 from both sides: \[ \frac{1}{2}x < 2. \] Multiply both sides by 2: \[ x < 4. \] This inequality represents a vertical line at \[ x = 4. \] Since the inequality is strict (\(<\)), draw the vertical line dashed and shade the region to the left of the line. --- 9. For the inequality \[ \frac{2}{3}x - \frac{10}{3} > -4, \] first eliminate the denominators by multiplying the entire inequality by 3: \[ 2x - 10 > -12. \] Then add 10 to both sides: \[ 2x > -2. \] Divide by 2: \[ x > -1. \] This gives a vertical boundary line at \[ x = -1, \] drawn dashed because the inequality is strict, with the region to the right of the line being shaded.