## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Using $y=mx+b$ where $m$ is the slope, the slope of the first equation, \begin{array}{l}\require{cancel} 2x+3y=1 \\\\ 3y=-2x+1 \\\\ y=-\dfrac{2}{3}x+\dfrac{1}{3} \end{array} is $m_1= -\dfrac{2}{3} .$ Using $y=mx+b$ where $m$ is the slope, the slope of the second equation, \begin{array}{l}\require{cancel} 2x-3y=5 \\\\ -3y=-2x+5 \\\\ y=\dfrac{-2}{-3}x+\dfrac{5}{-3} \\\\ y=\dfrac{2}{3}x-\dfrac{5}{3} \end{array} is $m_2= \dfrac{2}{3} .$ Since $m_1\ne m_2$ nor $m_1\cdot m_2\ne-1,$ then the given lines are $\text{ neither }$ parallel nor perpendicular.