Draw the boundary line for the inequality 2x+2y > 6, which has the equation the same as the inequality with the > replaced by an =. 2x+2y = 6 has x-intercept (3,0) and y-intercept (0,3) Draw it dotted because the inequality is > and not ≥.Test to see if the origin (0,0) is a solution. If it is, then all the solution points are on the same side of the line as the origin is on. Otherwise they are all on the side the origin is NOT on. We substitute the origin (x,y) = (0,0) in 2x+2y > 6 2(0)+2(0) > 6 0 > 6 That's false, so we know the solutions are all on the side the origin is NOT on, so we know that we'll be shading above and to the right of the line, for that's the side that the origin is NOT on. Next we draw the boundary line for the inequality x < 5, which has the equation the same as the inequality with the < replaced by an =. That's x = 5 which is a vertical line through 5 on the x-axis. Draw it dotted because the inequality is < and not ≤. Test to see if the origin (0,0) is a solution. If it is, then all the solution points are on the same side of the line as the origin is on. Otherwise they are all on the side the origin is NOT on. We substitute the origin (x,y) = (0,0) in x < 5 We only need to substitute 0 for x since there is no y in the equation: 0 < 5 That's true, so we know the solutions are all on the side the origin IS on, so we know that we'll be shading to the left of the vertical line, for that's the side that the origin IS on. Next we draw the boundary line for the inequality y ≥ 5, which has the equation the same as the inequality with the ≥ replaced by an =. That's y = -2 which is a horizontal line through -2 on the y-axis. Draw it solid because the inequality is ≥ and not >. Test to see if the origin (0,0) is a solution. If it is, then all the solution points are on the same side of the line as the origin is on. Otherwise they are all on the side the origin is NOT on. We substitute the origin (x,y) = (0,0) in y ≥ -2 We only need to substitute 0 for y since there is no x in the equation: 0 ≥ -2 That's true, so we know the solutions are all on the side the origin IS on, so we know that we'll be shading above the horizontal line, for that's the side that the origin IS on. Actually this line does not change the solution since we shade above it anyway. So we shade above the slanted line, to the left of the vertical line, which is of course above the horizontal line, so the finished graph is: Edwin