This is “Solving Inequalities with Two Variables”, section 2.7 from the book Advanced Algebra (v. 1.0). For details on it (including licensing), click here.
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We know that a linear equation with two variables has infinitely many ordered pair solutions that form a line when graphed. A linear inequality with two variablesAn inequality relating linear expressions with two variables. The solution set is a region defining half of the plane., on the other hand, has a solution set consisting of a region that defines half of the plane.
Linear Equation 
Linear Inequality 

$y=\frac{3}{2}x+3$ 
$y\le \frac{3}{2}x+3$ 


For the inequality, the line defines the boundary of the region that is shaded. This indicates that any ordered pair in the shaded region, including the boundary line, will satisfy the inequality. To see that this is the case, choose a few test pointsA point not on the boundary of the linear inequality used as a means to determine in which halfplane the solutions lie. and substitute them into the inequality.
Test point 
$y\le \frac{3}{2}x+3$ 
(0, 0) 
$\begin{array}{l}0\le \frac{3}{2}\left(0\right)+3\\ 0\le 3\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\text{\u2713}}\end{array}$ 
(2, 1) 
$\begin{array}{l}1\le \frac{3}{2}\left(2\right)+3\\ 1\le 3+3\\ 1\le 6\hspace{0.17em}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\text{\u2713}}\end{array}$ 
(−2, −1) 
$\begin{array}{l}1\le \frac{3}{2}\left(2\right)+3\\ 1\le 3+3\\ 1\le 0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\text{\u2713}}\end{array}$ 
Also, we can see that ordered pairs outside the shaded region do not solve the linear inequality.
Test point 
$y\le \frac{3}{2}x+3$ 
(−2, 3) 
$\begin{array}{l}3\le \frac{3}{2}\left(2\right)+3\\ 3\le 3+3\\ 3\le 0\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\text{\u2717}}\end{array}$ 
The graph of the solution set to a linear inequality is always a region. However, the boundary may not always be included in that set. In the previous example, the line was part of the solution set because of the “or equal to” part of the inclusive inequality $\le .$ If given a strict inequality $<$, we would then use a dashed line to indicate that those points are not included in the solution set.
NonInclusive Boundary 
Inclusive Boundary 

$y<\frac{3}{2}x+3$ 
$y\le \frac{3}{2}x+3$ 


Consider the point (0, 3) on the boundary; this ordered pair satisfies the linear equation. It is the “or equal to” part of the inclusive inequality that makes the ordered pair part of the solution set.
$y<\frac{3}{2}x+3$ 
$y\le \frac{3}{2}x+3$ 
$\begin{array}{l}3<\frac{3}{2}\left(0\right)+3\\ 3<0+3\\ 3<3\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\text{\hspace{0.17em}}{\text{\u2717}}\end{array}$ 
$\begin{array}{l}3\le \frac{3}{2}\left(0\right)+3\\ 3\le 0+3\\ 3\le 3\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\text{\u2713}}\end{array}$ 
So far we have seen examples of inequalities that were “less than.” Now consider the following graphs with the same boundary:
Greater Than (Above) 
Less Than (Below) 

$y\ge \frac{3}{2}x+3$ 
$y\le \frac{3}{2}x+3$ 


Given the graphs above, what might we expect if we use the origin (0, 0) as a test point?
$y\ge \frac{3}{2}x+3$ 
$y\le \frac{3}{2}x+3$ 
$\begin{array}{l}0\ge \frac{3}{2}\left(0\right)+3\\ 0\ge 0+3\\ 0\ge 3\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\text{\u2717}}\end{array}$ 
$\begin{array}{l}0\le \frac{3}{2}\left(0\right)+3\\ 0\le 0+3\\ 0\le 3\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\text{\u2713}}\end{array}$ 
Determine whether or not $\left(2,\frac{1}{2}\right)$ is a solution to $5x2y<10.$
Solution:
Substitute the x and yvalues into the equation and see if a true statement is obtained.
$$\begin{array}{ccc}\hfill 5x2y& <& 10\hfill \\ \hfill 5\left({2}\right)2\left({\frac{1}{2}}\right)& <& 10\hfill \\ \hfill 101& <& 10\hfill \\ \hfill 9& <& 10\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\text{\u2713}}\hfill \end{array}$$
Answer: $\left(2,\frac{1}{2}\right)$ is a solution.
These ideas and techniques extend to nonlinear inequalities with two variables. For example, all of the solutions to $y>{x}^{2}$ are shaded in the graph below.
The boundary of the region is a parabola, shown as a dashed curve on the graph, and is not part of the solution set. However, from the graph we expect the ordered pair (−1,4) to be a solution. Furthermore, we expect that ordered pairs that are not in the shaded region, such as (−3, 2), will not satisfy the inequality.
Check (−1,4) 
Check (−3, 2) 

$\begin{array}{l}y>{x}^{2}\\ 4>{\left(1\right)}^{2}\\ 4>1\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\text{\u2713}}\end{array}$ 
$\begin{array}{l}y>{x}^{2}\\ 2>{\left(3\right)}^{2}\\ 2>9\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\text{\u2717}}\end{array}$ 
Following are graphs of solutions sets of inequalities with inclusive parabolic boundaries.
$y\le {\left(x1\right)}^{2}2$ 
$y\ge {\left(x1\right)}^{2}2$ 


You are encouraged to test points in and out of each solution set that is graphed above.
Solutions to linear inequalities are a shaded halfplane, bounded by a solid line or a dashed line. This boundary is either included in the solution or not, depending on the given inequality. If we are given a strict inequality, we use a dashed line to indicate that the boundary is not included. If we are given an inclusive inequality, we use a solid line to indicate that it is included. The steps for graphing the solution set for an inequality with two variables are shown in the following example.
Graph the solution set $y>3x+1.$
Solution:
Step 1: Graph the boundary. Because of the strict inequality, we will graph the boundary $y=3x+1$ using a dashed line. We can see that the slope is $m=3=\frac{3}{1}=\frac{rise}{run}$ and the yintercept is (0, 1).
Step 2: Test a point that is not on the boundary. A common test point is the origin, (0, 0). The test point helps us determine which half of the plane to shade.
Test point 
$y>3x+1$ 
(0, 0) 
$\begin{array}{l}0>3(0)+1\\ 0>1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{\u2717}}\end{array}$ 
Answer:
Consider the problem of shading above or below the boundary line when the inequality is in slopeintercept form. If $y>mx+b$, then shade above the line. If $y<mx+b$, then shade below the line. Shade with caution; sometimes the boundary is given in standard form, in which case these rules do not apply.
Graph the solution set $2x5y\ge 10.$
Solution:
Here the boundary is defined by the line $2x5y=10.$ Since the inequality is inclusive, we graph the boundary using a solid line. In this case, graph the boundary line using intercepts.
To find the xintercept, set y = 0. 
To find the yintercept, set x = 0. 

$2x5y=10$ $\begin{array}{ccc}\hfill 2x5\left({0}\right)& =& 10\hfill \\ \hfill 2x& =& 10\hfill \\ \hfill x& =& 5\hfill \end{array}$ 
$2x5y=10$ $\begin{array}{ccc}\hfill 2\left({0}\right)5y& =& 10\hfill \\ \hfill 5y& =& 10\hfill \\ \hfill y& =& 2\hfill \end{array}$ 
xintercept: (−5, 0) 
yintercept: (0, 2) 
Next, test a point; this helps decide which region to shade.
Test point 
$2x5y\ge 10$ 
(0, 0) 
$\begin{array}{ccc}\hfill 2\left(0\right)5\left(0\right)& \ge & 10\hfill \\ \hfill 0& \ge & 10\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\text{\u2713}}\hfill \end{array}$ 
Since the test point is in the solution set, shade the half of the plane that contains it.
Answer:
In this example, notice that the solution set consists of all the ordered pairs below the boundary line. This may seem counterintuitive because the original inequality involved “greater than” $\ge .$ This illustrates that it is a best practice to actually test a point. Solve for y and you see that the shading is correct.
$$\begin{array}{ccc}\hfill 2x5y& \ge & 10\hfill \\ \hfill 2x5y{2x}& \ge & 10{2x}\hfill \\ \hfill 5y& \ge & 2x10\hfill \\ \hfill \frac{5y}{{5}}& {\le}& \frac{2x10}{{5}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{Reverse\text{\hspace{0.17em}}the\text{\hspace{0.17em}}inequality}.\hfill \\ \hfill y& \le & \frac{2}{5}x+2\hfill \end{array}$$
In slopeintercept form, you can see that the region below the boundary line should be shaded. An alternate approach is to first express the boundary in slopeintercept form, graph it, and then shade the appropriate region.
Graph the solution set $y<2.$
Solution:
First, graph the boundary line $y=2$ with a dashed line because of the strict inequality. Next, test a point.
Test point 
$y<2$ 
(0, 0) 
$0<2\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\text{\u2713}}$ 
In this case, shade the region that contains the test point.
Answer:
The steps are the same for nonlinear inequalities with two variables. Graph the boundary first and then test a point to determine which region contains the solutions.
Graph the solution set $y<{\left(x+2\right)}^{2}1.$
Solution:
The boundary is a basic parabola shifted 2 units to the left and 1 unit down. Begin by drawing a dashed parabolic boundary because of the strict inequality.
Next, test a point.
Test point 
$y<{\left(x+2\right)}^{2}1$ 
(0, 0) 
$\begin{array}{l}0<{\left(0+2\right)}^{2}1\\ 0<41\\ 0<3\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{\u2713}}\end{array}$ 
In this case, shade the region that contains the test point $\left(0,0\right).$
Answer:
Graph the solution set $y\ge {x}^{2}+3.$
Solution:
The boundary is a basic parabola shifted 3 units up. It is graphed using a solid curve because of the inclusive inequality.
Next, test a point.
Test point 
$y\ge {x}^{2}+3$ 
(0, 0) 
$\begin{array}{l}0\ge {0}^{2}+3\\ 0\ge 3\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{\u2717}}\end{array}$ 
In this case, shade the region that does not contain the test point $\left(0,0\right).$
Answer:
Is the ordered pair a solution to the given inequality?
$5xy>2\text{;\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(3,4\right)$
$4xy<8\text{;\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(3,10\right)$
$y\le {x}^{2}1$; $\left(1,1\right)$
$y\ge {x}^{2}+3$; $\left(2,0\right)$
$y\ge {\left(x5\right)}^{2}+1$; $\left(3,4\right)$
$y\le 2{\left(x+1\right)}^{2}3$; $\left(1,2\right)$
$y>3\leftx\right$; $\left(4,3\right)$
$y<\leftx\right8$; $\left(5,7\right)$
$y>\left2x1\right3$; $\left(1,3\right)$
$y<\left3x2\right+2$; $\left(2,10\right)$
Graph the solution set.
$y<2x1$
$y>4x+1$
$y\ge \frac{2}{3}x+3$
$y\le \frac{4}{3}x3$
$2x+3y\le 18$
$5x+2y\le 8$
$6x5y>30$
$8x6y<24$
$3x4y<0$
$x3y>0$
$x+y\ge 0$
$xy\ge 0$
$y\le 2$
$y>3$
$x<2$
$x\ge 3$
$5x\le 4y12$
$4x\le 123y$
$4y+2<3x$
$8x<96y$
$5\ge 3x15y$
Write an inequality that describes all points in the upper halfplane above the xaxis.
Write an inequality that describes all points in the lower halfplane below the xaxis.
Write an inequality that describes all points in the halfplane left of the yaxis.
Write an inequality that describes all points in the halfplane right of the yaxis.
Write an inequality that describes all ordered pairs whose ycoordinate is at least k units.
Write an inequality that describes all ordered pairs whose xcoordinate is at most k units.
Graph the solution set.
$y\le {x}^{2}+3$
$y>{x}^{2}2$
$y\le {x}^{2}$
$y\ge {x}^{2}$
$y>{\left(x+1\right)}^{2}$
$y>{\left(x2\right)}^{2}$
$y\le {\left(x1\right)}^{2}+2$
$y\le {\left(x+3\right)}^{2}1$
$y<{x}^{2}+1$
$y>{\left(x2\right)}^{2}+1$
$y\ge \leftx\right2$
$y<\leftx\right+1$
$y<\leftx3\right$
$y\le \leftx+2\right$
$y>\leftx+1\right$
$y\le \leftx2\right$
$y\ge \leftx+3\right2$
$y\ge \leftx2\right1$
$y<\leftx+4\right+2$
$y>\leftx4\right1$
$y>{x}^{3}1$
$y\le {x}^{3}+2$
$y\le \sqrt{x}$
$y>\sqrt{x}1$
A rectangular pen is to be constructed with at most 200 feet of fencing. Write a linear inequality in terms of the length l and the width w. Sketch the graph of all possible solutions to this problem.
A company sells one product for $8 and another for $12. How many of each product must be sold so that revenues are at least $2,400? Let x represent the number of products sold at $8 and let y represent the number of products sold at $12. Write a linear inequality in terms of x and y and sketch the graph of all possible solutions.
No
Yes
Yes
No
No
No
Yes
$y>0$
$x<0$
$y\ge k$
$l+w\le 100$;