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we have the following:
[tex]\begin{gathered} y=3x-4 \\ y=\frac{2}{5}x+9 \end{gathered}[/tex]solving the system by means of substitution:
[tex]\begin{gathered} 3x-4=\frac{2}{5}x+9 \\ 5\cdot3x-5\cdot4=5\cdot\frac{2}{5}x+5\cdot9 \\ 15x-20=2x+45 \\ 15x-2x=45+20 \\ 13x=65 \\ x=\frac{65}{13} \\ x=5 \end{gathered}[/tex]now, for x
[tex]y=3\cdot5-4=15-4=11[/tex]The solution is (5, 11)