## Intermediate Algebra (6th Edition)

Changing the given inequality, $5x-3y\lt10 ,$ to equality and then isolating $y$ result to \begin{array}{l}\require{cancel} 5x-3y=10 \\\\ -3y=-5x+10 \\\\ y=\dfrac{-5}{-3}x+\dfrac{10}{-3} \\\\ y=\dfrac{5}{3}x-\dfrac{10}{3} .\end{array} Use the table of values below to graph this line. Since the inequality used is "$\lt$", use a broken line. Using the testpoint $( 0,0 )$, then \begin{array}{l}\require{cancel} 5(0)-3(0)\lt10 \\\\ 0\lt10 \text{ (TRUE)} .\end{array} Since the solution above ended with a $\text{ TRUE }$ statement, then the testpoint is $\text{ part }$ of the solution set.