We have the next system of equations:
[tex]\begin{gathered} y=3x+19\text{ (eq. 1)} \\ y=5x+33\text{ (eq. 2) } \end{gathered}[/tex]Substituting y = 3x + 19 into the second equation, and solving for x:
[tex]\begin{gathered} 3x+19=5x+33 \\ 3x+19-3x-33=5x+33-3x-33 \\ -14=2x \\ \frac{-14}{2}=\frac{2x}{2} \\ -7=x \end{gathered}[/tex]Substituting x = -7 into the first equation:
[tex]\begin{gathered} y=3(-7)+19 \\ y=-21+19 \\ y=-2 \end{gathered}[/tex]The solution is (-7, -2)
