Lesson 6: Graphing Systems of Linear Inequalities

New Concept

A system of linear inequalities involves multiple inequalities. You'll learn to find the solution—the set of ordered pairs $(x, y)$ that satisfies all of them—by graphing and identifying the common shaded region, a key skill for real-world problems.

What’s next

Next, you'll tackle practice cards to determine if an ordered pair is a solution, then move on to graphing systems with interactive examples.

Property

Two or more linear inequalities grouped together form a system of linear inequalities.

To solve a system of linear inequalities, we will find values of the variables that are solutions to both inequalities. We solve the system by using the graphs of each inequality and show the solution as a graph. We will find the region on the plane that contains all ordered pairs $(x, y)$ that make both inequalities true.

Examples

Determine whether the ordered pair is a solution to the system:

• Is $(2, 3)$ a solution? For the first inequality, $2 + 2(3) \leq 8$, which is $8 \leq 8$ (True). For the second, $3(2) - 3 > 5$, which is $3 > 5$ (False). Since one is false, $(2, 3)$ is not a solution.
• Is $(4, 1)$ a solution? For the first inequality, $4 + 2(1) \leq 8$, which is $6 \leq 8$ (True). For the second, $3(4) - 1 > 5$, which is $11 > 5$ (True). Since both are true, $(4, 1)$ is a solution.
• Is $(-1, -10)$ a solution? For the first inequality, $-1 + 2(-10) \leq 8$, which is $-21 \leq 8$ (True). For the second, $3(-1) - (-10) > 5$, which is $7 > 5$ (True). Since both are true, $(-1, -10)$ is a solution.

Property

Solutions of a system of linear inequalities are the values of the variables that make all the inequalities true.

The solution of a system of linear inequalities is shown as a shaded region in the $x$-$y$ coordinate system that includes all the points whose ordered pairs make the inequalities true.

Examples

Consider the system:

• The solution is the region where the shading for $y < x + 1$ and $y \geq -2x + 4$ overlap. Any point in this doubly-shaded area is a solution.
• The point $(3, 3)$ is in the overlapping region. Let's check: $3 < 3 + 1$ (True) and $3 \geq -2(3) + 4$, so $3 \geq -2$ (True). It is a solution.
• The point $(0, 0)$ is not in the overlapping region. Let's check: $0 < 0 + 1$ (True), but $0 \geq -2(0) + 4$, so $0 \geq 4$ (False). It is not a solution.

Property

To solve a system of linear inequalities by graphing:

1. Graph the first inequality. Graph the boundary line, using a solid line for $\leq$ or $\geq$ and a dashed line for $<$ or $>$. Shade the side of the boundary line where the inequality is true.
2. On the same grid, graph the second inequality.
3. The solution is the region where the shading overlaps.
4. Check by choosing a test point in the overlapping region to ensure it makes both inequalities true.

Examples

• Solve the system:
  $$\begin{cases} y > x - 2 \\ y \leq -x + 4 \end{cases}$$
  Graph a dashed line for $y = x - 2$ and shade above it. Graph a solid line for $y = -x + 4$ and shade below it. The overlapping region is the solution.
• Solve the system:
  $$\begin{cases} x + y < 5 \\ y > 1 \end{cases}$$
  Graph a dashed line for $x+y=5$ and shade below it. Graph a dashed horizontal line for $y=1$ and shade above it. The solution is the overlapping triangle-like region.
• Solve the system:
  $$\begin{cases} x < 3 \\ y \geq -2 \end{cases}$$
  Graph a dashed vertical line at $x=3$ and shade to the left. Graph a solid horizontal line at $y=-2$ and shade above. The solution is the top-left quadrant defined by these lines.

Explanation

This is a visual way to find the 'sweet spot' that satisfies all conditions. You draw each inequality's boundary and shade its solution area. The zone where all the shaded areas overlap is the final answer.

Property

A system of inequalities has no solution if there is no point in both shaded regions. When graphed, the solution regions for the individual inequalities do not overlap at all. This means there is no ordered pair $(x, y)$ that makes all inequalities in the system true.

Examples

• Solve the system:
  $$\begin{cases} y > x + 3 \\ y < x - 1 \end{cases}$$
  The shaded regions are for areas above the line $y=x+3$ and below the line $y=x-1$. Since the lines are parallel and the regions are separate, there is no overlap and no solution.
• Solve the system:
  $$\begin{cases} x > 2 \\ x < -1 \end{cases}$$
  Shading to the right of $x=2$ and to the left of $x=-1$ results in two separate regions that never intersect. There is no solution.
• Solve the system:
  $$\begin{cases} y \leq -2x + 1 \\ y \geq -2x + 5 \end{cases}$$
  The shaded regions are below $y=-2x+1$ and above $y=-2x+5$. The regions do not overlap, so the system has no solution.

Explanation

Imagine being told to stand north of a line and south of a parallel line that's 'below' it. You can't do both! Similarly, some systems have contradictory rules, so their shaded areas never intersect, meaning no solution exists.

Property

To solve applications using systems of inequalities:

1. Let variables represent the quantities in the situation.
2. Write a system of inequalities that models the conditions of the problem (e.g., budget limits, quantity requirements).
3. Graph the system. The solution is the overlapping region, which represents all feasible options.
4. Interpret the solution in the context of the problem by checking if specific points lie within the solution region.

Examples

A student is buying notebooks ($n$) and pens ($p$). Notebooks cost 3 dollars each and pens cost 1 dollar each. She must buy at least 5 items in total but cannot spend more than 15 dollars.

• a) Write a system of inequalities: The system is
  $$\begin{cases} n + p \geq 5 \\ 3n + p \leq 15 \end{cases}$$
  with $n \geq 0$ and $p \geq 0$.
• b) Could she buy 3 notebooks and 4 pens? The point $(3, 4)$ gives $3+4 \geq 5$ (True) and $3(3)+4 \leq 15$ (True). Yes, this is a valid option.
• c) Could she buy 5 notebooks and 2 pens? The point $(5, 2)$ gives $5+2 \geq 5$ (True) but $3(5)+2 \leq 15$, or $17 \leq 15$ (False). No, this option is too expensive.

Explanation

This is where math solves real-world puzzles with multiple constraints, like managing a budget while needing a minimum number of items. The graph's solution area shows every possible combination that works within the given rules.