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In our study of linear equations in two variables, we observed that all the solutions to the equation, and only the solutions to the equation, were located on the graph of the equation. We now wish to determine the location of the solutions to linear inequalities in two variables. Linear inequalities in two variables are inequalities of the forms:
$$\begin{array}{cc}ax+by\le c& ax+by\ge c\\ ax+by<c& ax+by>c\end{array}$$
$2x+3y\le 6$
All the solutions to the inequality $2x+3y\le 6$ lie in the shaded half-plane.
Point $A(1,\text{\hspace{0.17em}}-1)$ is a solution since
$\begin{array}{l}2x+3y\le 6\hfill \\ 2(1)+3(-1)\le 6?\hfill \\ 2-3\le 6?\hfill \\ -1\le 6.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{True}\hfill \end{array}$
Point $B(2,\text{\hspace{0.17em}}5)$ is not a solution since
$\begin{array}{l}2x+3y\le 6\hfill \\ 2(2)+3(5)\le 6?\hfill \\ 4+15\le 6?\hfill \\ 19\le 6.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{False}\hfill \end{array}$
The method of graphing linear inequalities in two variables is as follows:
Graph $3x-2y\ge -\text{\hspace{0.17em}}4$ .
1. Graph the boundary line. The inequality is $\ge $ so we’ll draw the line solid . Consider the inequality as an equation.
$x$ | $y$ | $\left(x,\text{\hspace{0.17em}}y\right)$ |
$\begin{array}{r}\hfill 0\\ \hfill \frac{-4}{3}\end{array}$ | $\begin{array}{c}2\\ \\ 0\end{array}$ | $\begin{array}{l}\left(0,\text{\hspace{0.17em}}2\right)\hfill \\ \left(\frac{-4}{3},\text{\hspace{0.17em}}0\right)\hfill \end{array}$ |
2. Choose a test point. The easiest one is $\left(0,\text{\hspace{0.17em}}0\right)$ . Substitute $\left(0,\text{\hspace{0.17em}}0\right)$ into the original inequality.
Graph $x+y-3<0$ .
1. Graph the boundary line:
$x+y-3=0$ . The inequality is
$<$ so we’ll draw the line
dotted .
2. Choose a test point, say $(0,\text{\hspace{0.17em}}0)$ .
Graph $y\le 2x$ .
$\begin{array}{l}y\le 2x\hfill \\ 0\le 2\left(0\right)?\hfill \\ 0\le 0.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{True}\hfill \end{array}$
Shade the half-plane containing $\left(0,\text{\hspace{0.17em}}0\right)$ . We can’t! $\left(0,\text{\hspace{0.17em}}0\right)$ is right on the line! Pick another test point, say $\left(1,\text{\hspace{0.17em}}6\right)$ .$\begin{array}{l}y\le 2x\hfill \\ 6\le 2\left(1\right)?\hfill \\ 6\le 2.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{False}\hfill \end{array}$
Shade the half-plane on the opposite side of the boundary line.Graph $y>2$ .
1. Graph the boundary line
$y=2$ . The inequality is
$>$ so we’ll draw the line
dotted .
2. We don’t really need a test point. Where is
$y>2?$
Above the line
$y=2!$ Any point above the line clearly has a
$y\text{-coordinate}$ greater than 2.
Solve the following inequalities by graphing.
Solve the inequalities by graphing.
$x+y\le 1$
$-x+5y-10<0$
$2x+5y-15\ge 0$
$-2x+4y>0$
(
[link] ) Supply the missing word. The geometric representation (picture) of the solutions to an equation is called the
( [link] ) Supply the denominator: $m=\frac{{y}_{2}-{y}_{1}}{?}$ .
$m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$
(
[link] ) Graph the equation
$y=-3x+2$ .
( [link] ) Write the equation of the line that has slope 4 and passes through the point $\left(-1,\text{\hspace{0.17em}}2\right)$ .
$y=4x+6$
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