The solution set to a system of inequalities is often shown on a coordinate plane, that shows the boundary curves or lines that correspond to each inequality in the system. The region that satisfies the system of inequalities is shaded to show the area that contains all points that satisfy the system.

A solid line indicates the points on a line or curve are included in the solution and a dashed line indicates the points on a line or curve are not included in the solution.

Using test points in different quadrants can help with determining which regions to shade.

The graph of y equals negative x squared plus 4 drawn using solid lines, and the graph of y equals x squared drawn using dashed lines in a four quadrant coordinate plane. The region above the boundary parabola y equals negative x squared plus 4 is shaded green. The region above the boundary parabola y equals x squared is shaded blue.

Two inequalities on the same graph \\ y \geq -x^2+4 and y>x^2

A four quadrant coordinate plane. The graph of y equals negative x squared plus 4 under the domain interval negative square root of 2 to positive square root of 2, including negative square root of 2 and square root of 2 is drawn using solid lines. The graph of y equals x squared under the domain interval negative infinity to negative square root of 2 union positive square root of 2 to positive infinity, including negative square root of 2 and square root of 2 is drawn using dashed lines. The region both above the two boundary parabolas is shaded.

Solution set to the system

The solution to a system of inequalities in a given context is viable if the solution makes sense in the context, and is non-viable if it does not make sense in the context.

Worked examples

Example 1

Determine which of the following ordered pairs satisfies the given inequality:

\begin{cases} y < (x-3)(x-5) \\ y \geq 3x+2 \end{cases}

\left(0, 0\right)

Approach

To test if the ordered pairs satisfy the system of inequalities, we want to substitute the values into each inequality and see if it satisifes them. It must satisfy both inequalities to satisfy the entire system.

Solution

Consider the ordered pair \left(0,0\right)

We will first test if it satisfies the inequality y < (x-3)(x-5)

\displaystyle y	\displaystyle <	\displaystyle \left(x-3\right)\left(x-5\right)	State the inequality
\displaystyle 0	\displaystyle <	\displaystyle \left(0-3\right)\left(0-5\right)	Substitute y=0, x=0
\displaystyle 0	\displaystyle <	\displaystyle 15	Simplify

This inequality is true, so the ordered pair \left(0,0\right) satisfies the first inequality. We can now test the second inequality.

\displaystyle y	\displaystyle \geq	\displaystyle 3x+2	State the inequality
\displaystyle 0	\displaystyle \geq	\displaystyle 3\left(0\right)+2	Substitute x=0, y=0
\displaystyle 0	\displaystyle \geq	\displaystyle 2	Simplify

This inequality is false, so the ordered pair \left(0,0\right) does not satisfy the second inequality.

If the point fails to satisfy either of the inequalities in the system then it fails to satisfy the system of inequalities.

Reflection

We can also test the point by graphing the system of inequalities and seeing if the point lies in the shaded region.

The graph of y equals left parenthesis x minus 3 right parenthesis left parenthesis x minus 5 right parenthesis drawn using dashed lines, and the graph of y equals 3 x plus 2 drawn using solid lines in a four quadrant coordinate plane. The region below the boundary parabola, and above the boundary line is shaded. Point (0, 0) is plotted.

\left(-4,2\right)

Solution

Consider the ordered pair \left(-4,2\right)

We will first test if it satisfies the inequality y < (x-3)(x-5)

\displaystyle y	\displaystyle <	\displaystyle \left(x-3\right)\left(x-5\right)	State the inequality
\displaystyle 2	\displaystyle <	\displaystyle \left(-4-3\right)\left(-4-5\right)	Substitute x=-4,y=2
\displaystyle 0	\displaystyle <	\displaystyle 63	Simplify

This inequality is true, so the ordered pair \left(-4,2\right) satisfies the first inequality. We can now test the second inequality.

\displaystyle y	\displaystyle \geq	\displaystyle 3x+2	State the inequality
\displaystyle 2	\displaystyle \geq	\displaystyle 3\left(-4\right)+2	Substitute x=-4,y=2
\displaystyle 0	\displaystyle \geq	\displaystyle -10	Simplify

This inequality is true, so the ordered pair \left(-4,2\right) satisfies the second inequality as well.

If the point satisfies both inequalities in the system, then it satisfies the system of inequalities.

The ordered pair \left(-4, 2\right) satisfies the given system of inequalities.

Reflection

We can also test the point by graphing the system of inequalities and seeing if the point lies in the shaded region.

Example 2

Graph the boundary functions and shade the region which represents the solution set.

\begin{cases} y > x^2-2 \\ y \leq -2x^2-x+1 \end{cases}

Approach

To sketch the system of inequalities we can first construct the boundary curves for each inequality, namely y=x^2-2 and y=-2x^2-x+1. When given a strict inequality we will draw a dashed curve. When given a nonstrict inequality we will draw a solid curve.

The graph of y equals x squared minus 2 drawn using purple dashed lines, and the graph of y equals negative 2 x squared minus x plus 1 drawn using green solid lines in a four quadrant coordinate plane.

To determine which region will be shaded, we can choose some test points that satisfy each inequality individually, but lie on either side of the other inequality boundary. We can then shade the region that satisfies both inequalities.

For y > x^2-2, we know that as y must be greater than the boundary curve, the area above the curve satisfies the inequality. We can plot, for example, (0, 0) and (0,3).

For y \leq-2x^2-x+1, we know that as y must be less than or equal to the boundary curve, the area on or below the boundary curve satisfies the inequality. We can plot, for example, (0, -1) and (0,-4).

Solution

The region that will be shaded is the region which satisfies both inequalities. Using the test points we can see that the shading will occur in the region that is above the curve for \\y >x^2-2 but also below the curve for y\leq-2x^2-x+1.

Reflection

Another approach we could use to solve this particular problem is to test a point in each of the 5 distint regions that we can see formed by the intersection of these two graphs.

The point \left(0,0\right) lies in the middle region bounded by the two curves, and we can test if this point satisfies both inequalities.

\displaystyle y	\displaystyle >	\displaystyle x^2-2	State the inequality
\displaystyle 0	\displaystyle >	\displaystyle 0^2-2	Substitute x=0, y=0
\displaystyle 0	\displaystyle >	\displaystyle -2	Simplify
\displaystyle y	\displaystyle \leq	\displaystyle -2x^2-x+1	State the inequality
\displaystyle 0	\displaystyle \leq	\displaystyle -2\left(0^2\right)-0+1	Substitutex=0, y=0
\displaystyle 0	\displaystyle \leq	\displaystyle 1	Simplify

We can see that this point satisfies both inequalities, so the region that contains it must be the solution set for the system of inequalities.

Example 3

Write the system of inequalities that would determine the shaded region for the given graph.

A dashed boundary parabola and a solid boundary line plotted in a four quadrant coordinate plane. The dashed boundary parabola opens upward, and passes through points (5, negative 2), (3, 2), and (7, 2). The solid boundary line passes through (0, 2), and (negative 4, 0). The region below the dashed boundary parabola, and below the solid boundary line is shaded.

Approach

We want to identify the equation of the boundary line and the parabola that make up this system of inequalities. We can see that the linear inequality will be less than or equal to the values on the line, as it is a solid line, and that that the quadratic inequality will be less than the values on the parabola as it is a dashed line.

Solution

Since the parabola has a vertex at \left(5, -2\right) , we know that the vertex form of the equation will be:y=a\left(x-5\right)^2-2

for some value of a. We can find a by substituting in the coordinates of another point on the parabola such as \left(3, 2\right).

To find a:

\displaystyle y	\displaystyle =	\displaystyle a\left(x-5\right)^2-2	Vertex form
\displaystyle 2	\displaystyle =	\displaystyle a\left(3-5\right)^2-2	Substitute x=3, y=2
\displaystyle 2	\displaystyle =	\displaystyle 4a-2	Simplify
\displaystyle 4	\displaystyle =	\displaystyle 4a	Add 2
\displaystyle 1	\displaystyle =	\displaystyle a	Divide both sides by 4

The equation of the quadratic function in vertex form:y=\left(x-5\right)^2-2As the parabola is dashed, and the region below it is shaded the inequality is:y<\left(x-5\right)^2-2

The line has a y-intercept at \left(0, 2\right) and a slope of \dfrac{1}{2} and so has an equation in slope-intercept form of: y=\dfrac{x}{2} +2As the line is solid, and the region below it is shaded the inequality is:y\leq\dfrac{x}{2} +2We can now put them together to get our system of inequalities\begin{cases} y<\left(x-5\right)^2-2 \\ y\leq\dfrac{x}{2} +2 \end{cases}

Reflection

We can test our solution by choosing a point that lies in the shaded region, such as the origin \left(0,0 \right) and make sure it satisfies both inequalities.

We will first test if it satisfies the inequality y < \left(x-5\right)^2-2

\displaystyle y	\displaystyle <	\displaystyle \left(x-5\right)^2-2	State the inequality
\displaystyle 0	\displaystyle <	\displaystyle \left(0-5\right)^2-2	Substitute y=0, x=0
\displaystyle 0	\displaystyle <	\displaystyle 23	Simplify

This inequality is true, so the ordered pair \left(0,0\right) satisfies the first inequality. We can now test the second inequality.

\displaystyle y	\displaystyle \leq	\displaystyle \frac{x}{2}+2	State the inequality
\displaystyle 0	\displaystyle \leq	\displaystyle \frac{0}{2}+2	Substitute x=0, y=0
\displaystyle 0	\displaystyle \leq	\displaystyle 2	Simplify

This inequality is true, so the ordered pair \left(0,0\right) satisfies the second inequality also.

As the point satisfies both inequalities it satisfies the system of inequalities.

Outcomes

MA.912.AR.3.9

Given a mathematical or real-world context, write two-variable quadratic inequalities to represent relationships between quantities from a graph or a written description.

MA.912.AR.3.10

Given a mathematical or real-world context, graph the solution set to a two-variable quadratic inequality.

MA.912.AR.9.2

Given a mathematical or real-world context, solve a system consisting of a two-variable linear equation and a non-linear equation algebraically or graphically.

MA.912.AR.9.3

Given a mathematical or real-world context, solve a system consisting of two-variable linear or non-linear equations algebraically or graphically.

MA.912.AR.9.5

Graph the solution set of a system of two-variable inequalities.

MA.912.AR.9.7

Given a real-world context, represent constraints as systems of linear and non-linear equations or inequalities. Interpret solutions to problems as viable or non-viable options.

2.07 Nonlinear systems of inequalities

Concept summary

Worked examples

Example 1

Approach

Solution

Reflection

Solution

Reflection

Example 2

Approach

Solution

Reflection

Example 3

Approach

Solution

Reflection

Outcomes

MA.912.AR.3.9

MA.912.AR.3.10

MA.912.AR.9.2

MA.912.AR.9.3

MA.912.AR.9.5

MA.912.AR.9.7

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