System
[tex]\begin{gathered} 2x-5y=6 \\ y=3x+4 \end{gathered}[/tex]
Let's substitute Equation 2 into Equation 1 and solve for x
[tex]\begin{gathered} 2x-5(3x+4)=6 \\ 2x-5\cdot3x-5\cdot4=6 \\ 2x-15x-20=6 \\ -13x=6+20 \\ -13x=26 \\ x=\frac{26}{-13} \\ x=-2 \\ \end{gathered}[/tex]Now, solve for y
[tex]\begin{gathered} y=3x+4 \\ y=3\cdot(-2)+4 \\ y=-6+4 \\ y=-2 \\ \end{gathered}[/tex]The solution to the system is x = -2 and y = -2.
