Given :
[tex]{\qquad \sf \dashrightarrow y= 4x + 1 \: –––– \sf \: (i)}[/tex]
[tex]{\qquad \sf \dashrightarrow 3x + 2y = 13 \: –––– \sf \: (ii)}[/tex]
Now, Substituing the value of y in equation (ii) :
[tex]{\qquad \sf \dashrightarrow \: 3x + 2y = 13 }[/tex]
[tex]{\qquad \sf \dashrightarrow \: 3x + 2(4x + 1) = 13 }[/tex]
[tex]{\qquad \sf \dashrightarrow \: 3x + 8x + 2 = 13 }[/tex]
[tex]{\qquad \sf \dashrightarrow \: 11x + 2 = 13 }[/tex]
[tex]{\qquad \sf \dashrightarrow \: 11x = 13 - 2 }[/tex]
[tex]{\qquad \sf \dashrightarrow \: 11x = 11 }[/tex]
[tex]{\qquad \sf \dashrightarrow \: x = \dfrac{11}{11} }[/tex]
[tex]{\qquad \sf \dashrightarrow \bf \: x = 1 }[/tex]
Now, substituting the value of x in equation (i) :
[tex]{\qquad \sf \dashrightarrow y= 4x + 1}[/tex]
[tex]{\qquad \sf \dashrightarrow y= 4(1) + 1}[/tex]
[tex]{\qquad \sf \dashrightarrow y= 4 + 1}[/tex]
[tex]{\qquad \bf \dashrightarrow y= 5}[/tex]
