Question

Graph the solution set of the following linear inequality:3 - 3y > 21 - 3xAnswerKeypadKeyboard ShortcutsThe line will be drawn once all required data is provided and will update whenever a value is updated. Theregions will be added once the line is drawn.Choose the type of boundary line:Dashed -O Solid (-) OsEnter two points on the boundary line:10-5$-5Select the region you wish to be shaded:ОАOB101

Answer 1

Step-by-step explanation

To graph an inequality, treat the <, ≤, >, or ≥ sign as an = sign, and graph the equation. If the inequality is < or >, graph the equation as a dotted line. If the inequality is ≤ or ≥, graph the equation as a solid line

`3-3y>21-3x`

Step 1

a)change the "> " for a "=" and then isolate y

`\begin{gathered} 3-3y>21-3x \\ \text{swap the sign} \\ 3-3y=21-3x \\ \text{subtract 3 in both sides} \\ 3-3y-3=21-3x-3 \\ -3y=-3x+18 \\ \text{divide both sides by -3} \\ (-3y)/(-3)=(-3x+18)/(-3) \\ y=(-3x)/(-3)+(18)/(-3) \\ y=x-6 \end{gathered}`

b) now, graph the line

to do that, we need points

so

I) for x=0

`\begin{gathered} y=x-6 \\ y=0-6 \\ y=-6 \end{gathered}`

so,the coordinate is ( 0,-6)

ii) when x = 6

`\begin{gathered} y=6-6 \\ y=6-6 \\ y=0 \end{gathered}`

so, the coordinate is (6,0)

c) finally, draw a dotted line that passes trought (0,-6) and (6,0), as we are lookinjg for the y-values greater than the line, let's isolate y in the inequality to check what area to shade, so

[tex]\begin{gathered} 3-3y>21-3x \\ \text{subtract 3 in both sides} \\ 3-3y-3>21-3x-3 \\ -3y>-3x+18 \\ \text{divide both sides by (-3),remember swap the sign } \\ \frac{-3y}{-3}<\frac{-3x+18}{-3} \\ yso, we need to find all the values smaller than the line, so we need to shade de area under the line