3.4 Graphing Linear Inequalities and Systems

3.4 Graphing Linear Inequalities and Systems

Quick Answer: Graph $y > mx + b$ as the line $y = mx + b$ drawn dashed (the boundary is not included) and shade the side that a test point satisfies. Use a solid line for $\ge$ or $\le$. A system of two linear equations is solved graphically at the intersection point. A system of inequalities is the overlap of the shaded regions.

/practice/accuplacer-advanced-algebraPractice questions with detailed explanations

One Linear Inequality in Two Variables

Start with the boundary line $y = mx + b$ (or $Ax + By = C$). Then two decisions:

The reliable shade test is a test point not on the line, usually $(0, 0)$ unless the line through the origin makes $(0, 0)$ useless.

Worked example. Graph $y > -\frac{1}{2}x + 3$.

1. Graph $y = -\frac{1}{2}x + 3$: y-intercept $(0, 3)$, slope down 1, right 2. Draw it dashed.
2. Test $(0, 0)$: is $0 > -\frac{1}{2}(0) + 3$? Is $0 > 3$? No.
3. Shade the side that does not contain $(0, 0)$ — the side above the line, because $y$ is greater than the line's y-value.

When the inequality is already solved for $y$, a shortcut matches the inequality direction: $y > mx + b$ shades above; $y < mx + b$ shades below. When it is not solved for $y$, do not guess above. For $2x + 4y \le 8$, first write $y \le -\frac{1}{2}x + 2$ (dividing by a positive 4 keeps the inequality direction), then shade on and below a solid line. If you divide by a negative number, flip the inequality — the same rule as one-variable inequalities.

Worked example with a reversed inequality. $-3x + y < 6$ rearranges to $y < 3x + 6$. Dashed line through $(0, 6)$ with slope $3$; shade below. Equivalent check: $-2x - y \ge 4$ → multiply by $-1$ and flip → $2x + y \le -4$ → $y \le -2x - 4$. Solid line, shade below. Substituting $(0, 0)$ into $y \le -2x - 4$ gives $0 \le -4$, false, so $(0, 0)$ is not in the shade — consistent with shading below a line whose intercept is already at $-4$.

Vertical and horizontal inequalities follow the four slope cases from Section 3.1. $x > 2$ is a dashed vertical line at $x = 2$, shade to the right. $x \le -1$ is a solid vertical line at $x = -1$, shade to the left. $y \le -1$ is a solid horizontal line at $y = -1$, shade downward. $y > 4$ is a dashed horizontal line at $y = 4$, shade upward.

Systems of Two Linear Equations, Graphically

Two lines in the plane meet in one of three ways — the same three cases you solve algebraically in the linear-equations chapter:

Worked example. Solve graphically:

$y = 2x - 1$ $y = -x + 5$

The first line has slope $2$, intercept $-1$. The second has slope $-1$, intercept $5$. Different slopes ⇒ they intersect once. Set $2x - 1 = -x + 5$ to read the intersection exactly: $3x = 6$, $x = 2$, $y = 3$. The graphical solution is the point $(2, 3)$. Check both originals: $2(2) - 1 = 3$ and $-2 + 5 = 3$. On a CAT grid you would see the two lines cross at that lattice point; the algebraic confirmation is the same substitution or elimination you practiced at /study-guides/accuplacer-advanced-algebra/linear-equations/systems-of-linear-equations.

If the second equation had been $y = 2x + 4$, the graph would be two parallel lines (both slope 2, intercepts $-1$ and $4$) and the system would have no solution. If it had been $2y = 4x - 2$, it would simplify to $y = 2x - 1$, the same line, and every point on that line would be a solution.

Read a printed graph the same way: if the lines visibly cross, estimate the intersection, then check it in both equations. AAF will put the crossing slightly off a nearby lattice point that looks nice, to catch students who only glance. If the crossing appears near $(3, 2)$ but $(3, 2)$ fails one equation, compute the exact intersection instead of rounding to the pretty point.

Systems of Linear Inequalities

A system of inequalities asks for the region that satisfies every inequality at once — the overlap (intersection) of the individual shaded half-planes.

Worked example. Graph the solution set of

$y \ge x - 1$ $y < -2x + 4$

1. Solid line $y = x - 1$, shade above (test $(0, 0)$: $0 \ge -1$, true, so include the origin side).
2. Dashed line $y = -2x + 4$, shade below (test $(0, 0)$: $0 < 4$, true).
3. The solution is the wedge where those two shadings overlap. The boundary of the first line is included; the boundary of the second is not. Their intersection point — $x - 1 = -2x + 4$, $3x = 5$, $x = 5/3$, $y = 2/3$ — sits on the solid line and off the dashed line, so it is not in the solution set.

A second test point inside the intended wedge, such as $(0, 2)$: $2 \ge 0 - 1$ is true and $2 < 4$ is true, so $(0, 2)$ is in. A point such as $(3, 0)$: $0 \ge 3 - 1$ is false, so it is out even though it is below the dashed line.

For three inequalities the overlap can be a triangle (a bounded feasible region). AAF typically stays at two inequalities, but the test-point logic does not change: a point is in the solution only if it satisfies all of them.

Connecting the Picture Back to Algebra

Graphical and algebraic views of a system are the same object:

• Intersection point $\leftrightarrow$ unique solution of the two equations.
• Parallel distinct lines $\leftrightarrow$ $0 = $ a nonzero number, no solution.
• Coincident lines $\leftrightarrow$ $0 = 0$, infinitely many solutions.
• Shaded overlap $\leftrightarrow$ the set of ordered pairs that satisfy every inequality.

When an item shows a shaded graph and asks which system it matches, recover each boundary's equation from intercepts or slope, then use a test point inside the shade to choose $>$ versus $<$. That is the reverse of the drawing process, and it is a favorite read-the-graph task in this content area.

Keep practicing both directions — equation to graph and graph to equation. Use this chapter's slope and intercept skills to draw the boundaries, and use the linear-equations chapter's substitution and elimination to confirm intersection points exactly.