Systems of Inequalities: Practice

A point is a solution when it makes the inequality true, so add each point's coordinates and compare with $5$.

$2 + 2 = 4 \le 5 \text{ is true}$

The other points give sums $7$, $6$, and $7$, all greater than $5$, so only $(2, 2)$ satisfies the inequality.

Check whether each point's $y$ is strictly greater than $x + 1$.

$3 > 0 + 1 = 1 \text{ is true}$

For $(2, 2)$ you need $2 > 3$ (false), for $(1, 2)$ you need $2 > 2$ (false, not strict), and for $(3, 1)$ you need $1 > 4$ (false). Only $(0, 3)$ works.

The boundary is drawn solid when the inequality allows equality and dashed when it does not.

$\le \text{ includes the points where } 2x + y = 8$

Because $\le$ counts the points on the line as solutions, the boundary is solid. A strict $<$ would be dashed.

A strict inequality excludes the points on the boundary line itself.

$> \text{ does not allow } y = 3x - 2$

Since the points where $y = 3x - 2$ are not solutions, the boundary is drawn dashed.

Substitute the origin into the inequality.

$0 + 0 = 0, \quad 0 < 4 \text{ is true}$

The statement is true, so $(0, 0)$ is a solution, and the whole side of the line containing the origin is shaded.

Substitute the origin into the inequality.

$2(0) - 0 = 0, \quad 0 \ge 5 \text{ is false}$

The statement is false, so $(0, 0)$ is not a solution. You would shade the side of the line that does not contain the origin.

A point is on the line when its coordinates make $y = 2x - 1$ a true equation.

$2(2) - 1 = 3 = y$

So $(2, 3)$ is on the line. The others give $y = 1$, $y = -1$, and $y = 5$ for their $x$-values, none matching.

Substitute the point and compare the two sides.

$-x + 6 = -2 + 6 = 4, \quad 2 \le 4 \text{ is true}$

Since $2 \le 4$ holds, $(2, 2)$ is a solution.

The boundary line splits the plane into two sides, and one entire side satisfies the inequality.

$x - y \ge 1 \text{ holds on one side of } x - y = 1$

That side, together with the boundary when equality is allowed, is a half-plane.

A solution of a system must satisfy every inequality, so check both.

$2 + 1 = 3 \le 4 \quad\text{and}\quad 2 \ge 1$

Both statements are true, so $(2, 1)$ lies in the feasible region of the system.

The boundary is the equation you get by replacing the inequality symbol with an equals sign.

$y \ge 2 \;\rightarrow\; y = 2$

The graph of $y = 2$ is the horizontal line at height $2$, and the region $y \ge 2$ is everything on or above it.

The inequality $x < 3$ depends only on the $x$-coordinate, no matter how large $y$ is.

$1 < 3 \text{ is true}$

Only $(1, 100)$ has $x < 3$. The point $(3, 0)$ fails because $3 < 3$ is false, and the others have $x = 4$ and $x = 5$.