Q22. - Algebra 2

Question

Graph each inequality.

$x-6y+3>0$.

Step-by-Step Solution

Answer

The graph of the inequality is shown below:

Step 1 – Find the boundary line equation of the given equation.

Consider the inequality $x-6y+3>0$.

The boundary line equaton is $x-6y+3=0$.

Here the inequality is $>$, graph the boundary line as a dashed line.

Find the slope intercept form of $x-6y+3=0$:

$\begin{array}{l}x-6y+3=0\\ x-6y=-3\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}Subtract 3 from both sides}\\ -6y=-x-3\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}Subtract }x\text{ from both sides}\\ y=\frac{-x-3}{-6}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}Divide both sides by }-6\\ y=\frac{1}{6}x+\frac{1}{2}\end{array}$

The slope intercept form is $y=\frac{1}{6}x+\frac{1}{2}$with slope $\frac{1}{6}.$

Step 2 – Choose a test point and test the inequality.

Choose the test point as $\left(0,0\right).$

Substitute $x=0,y=0$in $x-6y+3>0$:

$\begin{array}{l}x-6y+3>0\\ 0-6\left(0\right)+3>0\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}substitute 0 for }x,0\text{ for }y\\ 3>0\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}true}\end{array}$

Step 3 – Conclusion

Since, $3>0$results a true inequality, so, shade the region that contains the point $\left(0,0\right).$

Graphical representation of $x-6y+3>0$: