## Intermediate Algebra (12th Edition)

A line in slope-intercept form has the equation: $y=mx+b$ ($m=slope$, $b=y-intercept$) We put our first equation into slope-intercept form: $3x-y\geq 3$ $-y\geq -3x+3$ $y\leq 3x-3$ Thus we graph a line with $slope=3$ and $y-intercept=-3$. We know that if an inequality has an "equal" sign (e.g. "$\leq$" or "$\geq$"), then a solid line should be graphed. If the inequality does not have an "equal" sign (e.g. "$\lt$" or "$\gt$"), then a dashed line should be graphed. We also know that if an inequality has a "less than" sign (e.g. "$y\lt$..." or "$y\leq$..." or "$x\lt$..." or "$x\leq$..."), then the shading should be below the line (or left for vertical lines). If the inequality has a "greater than" sign (e.g. "$y\gt$..." or "$y\geq$..." or "$x\gt$..." or "$x\geq$..."), then the shading should be above the line (or right for vertical lines). In our case, we must graph a solid line and shade below. Next, we look at our second equation: $y\lt 3$ This has the form of a horizontal line ($y=constant$) going through $y=3$. We see from the inequality that we must graph a dashed line and shade below. See the resulting graph. The dark region is the area where both inequalities overlap (e.g. the solution region).