Math Core

Real limits rarely say "exactly." A budget says at most $200; a plan needs at least $15$ hours. When two or more such conditions must hold together, you have a system of linear inequalities. Its solutions aren't a single point but a whole region of the plane, and graphing is the clearest way to find it.

Review: graphing one inequality

A linear inequality in two variables, such as $y > 2x - 1$ , is solved by every point on one side of the line $y = 2x - 1$ .

Graph the boundary line $y = 2x - 1$ . Draw it dashed for $<$ or $>$ (points on the line are not solutions) and solid for $\le$ or $\ge$ (they are).
Shade the correct side. When the inequality is solved for $y$ , $y >$ or $y \ge$ means shade above the line, and $y <$ or $y \le$ means shade below. Or use a test point not on the line, such as $(0, 0)$ : if it makes the inequality true, shade its side; if not, shade the other side.

Solving a system of inequalities

Definition

Solution of a system of inequalities

A solution of a system of inequalities is an ordered pair that makes every inequality in the system true. The set of all solutions is the region where the shaded regions of all the inequalities overlap.

Graph each inequality on the same plane. The overlap, where the shadings stack, is the solution region.

y > 2x − 1 (above the dashed line) and y ≤ −x + 3 (below the solid line). The darker overlap on the left is the solution region.Open in grapher →

In the picture, the dashed line shows that points on $y = 2x - 1$ are excluded, and the solid line shows that points on $y = -x + 3$ are included (as long as they also satisfy the other inequality).

Graphing a system of inequalities

Solve each inequality for $y$ if it isn't already.
Graph each boundary line: dashed for $<$ or $>$ , solid for $\le$ or $\ge$ .
Shade the correct side of each line.
The solution set is the region where all the shadings overlap.
Check a point from that region in every original inequality.

Worked example: Graphing and checking

Graph the system and decide whether $(0, 0)$ and $(2, 0)$ are solutions.

\begin{cases} y \ge x - 2 \\ y < -2x + 4 \end{cases}

Line $y = x - 2$ is solid (because of $\ge$ ); shade above it. Line $y = -2x + 4$ is dashed (because of $<$ ); shade below it.

The overlap lies above the solid line and below the dashed line. The boundary lines cross at (2, 0).Open in grapher →

Test $(0, 0)$ : $0 \ge 0 - 2$ is true, and $0 < 0 + 4$ is true. It's a solution, and it sits in the overlap.

Test $(2, 0)$ : this is where the boundaries cross. $0 \ge 2 - 2$ is true, but $0 < -4 + 4$ , or $0 < 0$ , is false. It lies on the dashed line, so it is not a solution, even though it's a corner of the region.

Inequalities in standard form

If an inequality is in standard form, solve it for $y$ first. Remember the rule from the inequalities unit: multiplying or dividing by a negative number reverses the inequality sign.

Worked example: Rewriting before graphing

Graph the system.

\begin{cases} 2x + 3y \le 12 \\ x - y > 1 \end{cases}

Solve each for $y$ :

\begin{aligned} 2x + 3y \le 12 \quad &\Longrightarrow \quad 3y \le -2x + 12 \quad \Longrightarrow \quad y \le -\tfrac{2}{3}x + 4 \\ x - y > 1 \quad &\Longrightarrow \quad -y > -x + 1 \quad \Longrightarrow \quad y < x - 1 \end{aligned}

In the second line, dividing by $-1$ flipped $>$ to $<$ . So shade below the solid line $y = -\tfrac{2}{3}x + 4$ and below the dashed line $y = x - 1$ .

Both regions lie below their lines. The overlap is the region below both lines. Its top corner is where the lines cross, at (3, 2).Open in grapher →

Check with $(4, 0)$ , which lies in the overlap: $2(4) + 3(0) = 8 \le 12$ is true, and $4 - 0 = 4 > 1$ is true.

Common mistake

Forgetting to flip the sign is the most common error here. From $x - y > 1$ , it's tempting to write $y > x - 1$ , which shades the wrong side. Protect yourself with a test point in the original inequality: $(0, 0)$ gives $0 - 0 > 1$ , which is false, so the solution region must not contain the origin.

When the regions don't overlap

A system of inequalities can have no solution. For example, $y > x + 3$ asks for points above one line, and $y < x - 1$ asks for points below a parallel line that sits $4$ units lower. No point is both above the higher line and below the lower one, so the shaded regions never meet and the system has no solution.

The two shaded regions are separated by a strip between the parallel lines. They never overlap.Open in grapher →

Modeling with systems of inequalities

In real situations the variables often count things, so they can't be negative. Constraints like $x \ge 0$ and $y \ge 0$ are part of the system and keep the region in the first quadrant.

Worked example: Earning a target

Maya babysits for $12 per hour and tutors for $20 per hour. This month she can work at most $15$ hours, and she wants to earn at least $200. Write a system, graph it, and decide whether $5$ hours of babysitting and $8$ hours of tutoring meets both goals.

Let $x$ be hours babysitting and $y$ hours tutoring.

\begin{cases} x + y \le 15 & \text{hours available} \\ 12x + 20y \ge 200 & \text{earnings goal} \\ x \ge 0, \; y \ge 0 & \text{hours can't be negative} \end{cases}

In slope-intercept form, the first two are $y \le -x + 15$ and $y \ge -0.6x + 10$ .

Maya's options: at or below y = −x + 15, at or above y = −0.6x + 10, in the first quadrant.Open in grapher →

Test $(5, 8)$ : $5 + 8 = 13 \le 15$ is true, and $12(5) + 20(8) = 60 + 160 = 220 \ge 200$ is true. Yes, that schedule works: $13$ hours for $220.

By contrast, $(10, 2)$ uses only $12$ hours but earns $120 + 40 = 160$ dollars, which is short of the goal. It lies below the earnings line, outside the region.

Tip

A point exactly on a solid boundary line counts as a solution (if it satisfies the other inequalities), and a point on a dashed line never does. When you check a boundary point, write out the comparison, like $0 < 0$ , rather than trusting the picture.

Practice

Practice 1

Is $(2, 1)$ a solution of the system $y < x + 1$ and $y \ge -x + 2$ ?

Practice 2

Which point is a solution of the system $y > 2x - 3$ and $y \le -x + 4$ ?

Practice 3

When you graph the system $y \le 3x + 1$ and $y > -x$ , which boundary lines are dashed?

Practice 4

Is $(1, 3)$ a solution of the system $y > 2x + 1$ and $y \le 5$ ?

Practice 5

Which inequality is equivalent to $x - 2y > 4$ ?

Practice 6

Which system has no solution?

Practice 7

A theater has $200$ seats. Adult tickets cost $10 and child tickets cost $6. The theater wants to sell at most $200$ tickets and bring in at least $1,500. If it sells $50$ child tickets, what is the least number of adult tickets that meets both conditions?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

How many points with whole-number coordinates satisfy the system $x \ge 1$ , $y \ge 1$ and $x + y \le 4$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.