Solutions to an inequality fill a shaded region; a dashed boundary is excluded

The shaded overlap satisfies both inequalities

In SAT Desmos, enter $y<2x-1$ and $y\ge -x+6$. The shared shaded region contains the points that satisfy both conditions. A strict inequality uses a dashed boundary because equality is excluded; an inclusive inequality uses a solid boundary.

Test $(4,4)$: $4<7$ and $4\ge2$, so it belongs to the overlap. Test $(2,4)$: $4<3$ is false, so it does not. Substitution is a fast check when the shading is difficult to read.

$\displaystyle y<2x-1$

Below the boundary; points on the line are excluded.

$\displaystyle y\ge -x+6$

Above the boundary; points on the line are included.

The point where boundary lines cross can fail a strict inequality

The two boundary lines meet at $(7/3,11/3)$. That point satisfies both boundary equations but fails $y<2x-1$, because it gives equality. A boundary intersection is not automatically a solution to the inequalities.

For a one-variable inequality such as $(x-2)(x+4)>0$, the zeros $-4$ and $2$ split the number line into intervals. Test a point in each interval. The solution is $x<-4$ or $x>2$; the strict inequality excludes the endpoints.

Restrict inputs with braces when the context limits them

In SAT Desmos, append braces to restrict a graph, for example $y=-x^2+6x\{0\le x\le6\}$. A time variable may require $x\ge0$, and a count may require whole numbers. A graphing window only changes what you see; it does not impose a domain.

Use the point lesson below to practice ordered pairs and substitution. You can also enter the inequalities and brace restrictions above in its calculator.

Keep practicing the connected skills