Exercise 45 Page 262

Graphing $\text{-}4x+2y>6$

To begin, we will isolate $y.$ This will make graphing the boundary line easier. Let's write the boundary line. To do so, we substitute the inequality sign with an equals sign. The boundary line has a slope of $2$ and a $y\text{-}$intercept of $3.$ Notice that we have a strict inequality. Therefore, the boundary line will be dashed. To decide which side to shade, we can use a test point that is not on the line. If the inequality holds true when substituting the coordinates of this point, we will shade the region that contains it. Otherwise, we will shade the opposite region. For simplicity, let's use $(\text{-}2,2).$ The inequality holds true. This means we should shade the side of the boundary line that contains the point.

Writing the Second Inequality

For our second inequality not to overlap with the one above, the boundary lines must be parallel. Otherwise, the lines will inevitably intersect and the solution sets will overlap. Moreover, to also avoid overlapping solution sets, the $y\text{-}$intercept of our line must be less than $3.$ Let's arbitrarily choose a $y\text{-}$intercept of $\text{-}1.$ Let's draw this line on the same coordinate plane. To shade the region below the second boundary line, we have to determine the inequality sign. To do so, we will use the point $(2,0).$ When substituted into the inequality, it must produce a true statement. Note that we do not really care if the line is dashed or solid. For simplicity, we will leave it solid and the inequality will be not strict. The inequality sign is less than or equal to. Therefore, our inequality is $y\le 2x-1.$ Let's shade the region! Finally, we form our system of inequalities by combining the inequalities.