Graphing a single inequality involves two main steps.

Plotting the boundary line.
Shading half of the plane to show the solution set.

For this exercise, we need to do this process for each of the inequalities in the system.

{6 x - 5 y < 15 x + 2 y \geq 7 (I) (II)

The system's solution set will be the intersection of the shaded regions in the graphs of Inequalities (I) and (II).

Boundary Lines

We can tell a lot of information about the boundary lines from the inequalities given in the system.

Exchanging the $inequality$ $symbols$ for $equals$ $signs$ gives us the boundary line equations.
Observing the $inequality$ $symbols$ tells us whether the inequalities are strict.
Writing the equations in slope-intercept form will help us highlight the slopes $m$ and $y -$ intercepts $b$ of the boundary lines.

Let's find each of these key pieces of information for the inequalities in the system.

Information	Inequality (I)	Inequality (II)
Given Inequality	$6 x - 5 y < 15$	$x + 2 y \geq 7$
Boundary Line Equation	$6 x - 5 y = 15$	$x + 2 y = 7$
Solid or Dashed?	$< \Rightarrow$ Dashed	$\geq \Rightarrow$ Solid
$y = m x + b$	$y = \frac{6}{5} x + (- 3)$	$y = - \frac{1}{2} x + \frac{7}{2}$

Great! With all of this information, we can plot the boundary lines.

Shading the Solution Sets

Before we can shade the solution set for each inequality, we need to determine on which side of the plane their solution sets lie. To do that, we will need a test point that does not lie on either boundary line.

We will use $(0, 0)$ as a test point. To do so, we will substitute $0$ for $x$ and $y$ in the given inequalities and simplify. If the substitution creates a true statement, we will shade the region that contains the point $(0, 0) .$ Otherwise, we will shade the region that does not contain the point.

Information	Inequality (I)	Inequality (II)
Given Inequality	$6 x - 5 y < 15$	$x + 2 y \geq 7$
Substitute $(0, 0)$	$6 (0) - 5 (0) < ? 15$	$0 + 2 (0) \geq ? 7$
Simplify	$0 < 15$	$0 ≱ 7$
Shaded Region	contains $(0, 0)$	does not contain $(0, 0)$

For Inequality (I), we will shade the region containing our test point, or above the boundary line. For Inequality (II), however, we will shade the region opposite the test point, or above the boundary line.

The solution set of the system is the intersection of the shaded regions.

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