{{ item.displayTitle }}

No history yet!

Student

Teacher

{{ item.displayTitle }}

{{ item.subject.displayTitle }}

{{ searchError }}

{{ courseTrack.displayTitle }} {{ statistics.percent }}% Sign in to view progress

{{ printedBook.courseTrack.name }} {{ printedBook.name }} Graphing a single inequality involves two main steps.

- Drawing the boundary line.
- Shading a half-plane to show the solution set.

For this exercise, we need to do this process for each of the inequalities in the system. ${4x+y>-1y<-4x+1 (I)(II) $
The system's solution set will be the *intersection* of the shaded regions in the graphs of Inequality (I) and Inequality (II).

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 lines in slope-intercept form will help us highlight each slope $m$ and $y-$intercept $b.$

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

Information | Inequality (I) | Inequality (II) |
---|---|---|

Given Inequality | $4x+y>-1$ | $y<-4x+1$ |

Boundary Line Equation | $4x+y=-1$ | $y=-4x+1$ |

Solid or Dashed? | $>⇒$ Dashed | $<⇒$ Dashed |

$y=mx+b$ | $y=-4x+(-1)$ | $y=-4x+1$ |

Great! With all of this information, we can draw the boundary lines. For each line, we start by plotting the $y-$intercept and using the slope to obtain a second point. Then, we connect those points with a straight edge.

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.

It looks like the point $(1,1)$ would be a good test point. We will substitute the coordinates of this point for $x$ and $y$ in the given inequalities and simplify. If the substitution creates a true statement, we shade the **same** region that contains the point. Otherwise, we shade the **opposite** region.

Information | Inequality (I) | Inequality (II) |
---|---|---|

Given Inequality | $4x+y>-1$ | $y<-4x+1$ |

Substitute $(1,1)$ | $4(1)+1>? -1$ | $1<? -4(1)+1$ |

Simplify | $5>-1$ | $1≮-3$ |

Shaded Region | same | opposite |

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 *below* the boundary line.

The overlapping portion of the inequalities is the solution set of the system of inequalities.