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Let [tex]X(s)[/tex] and [tex]Y(s)[/tex] denote the Laplace transforms of [tex]x(t)[/tex] and [tex]y(t)[/tex].
Taking the Laplace transform of both sides of both equations, we have
[tex]\dfrac{dx}{dt} + 3x + \dfrac{dy}{dt} = 1 \implies \left(sX(s) - x(0)\right) + 3X(s) + \left(sY(s) - y(0)\right) = \dfrac1s \\\\ \implies (s+3) X(s) + s Y(s) = \dfrac1s[/tex]
[tex]\dfrac{dx}{dt} - x + \dfrac{dy}{dt} = e^t \implies \left(sX(s) - x(0)\right) - X(s) + \left(sY(s) - y(0)\right) = \dfrac1{s-1} \\\\ \implies (s-1) X(s) + s Y(s) = \dfrac1{s-1}[/tex]
Eliminating [tex]Y(s)[/tex], we get
[tex]\left((s+3) X(s) + s Y(s)\right) - \left((s-1) X(s) + s Y(s)\right) = \dfrac1s - \dfrac1{s-1} \\\\ \implies X(s) = \dfrac14 \left(\dfrac1s - \dfrac1{s-1}\right)[/tex]
Take the inverse transform of both sides to solve for [tex]x(t)[/tex].
[tex]\boxed{x(t) = \dfrac14 (1 - e^t)}[/tex]
Solve for [tex]Y(s)[/tex].
[tex](s - 1) X(s) + s Y(s) = \dfrac1{s-1} \implies -\dfrac1{4s} + s Y(s) = \dfrac1{s-1} \\\\ \implies s Y(s) = \dfrac1{s-1} + \dfrac1{4s} \\\\ \implies Y(s) = \dfrac1{s(s-1)} + \dfrac1{4s^2} \\\\ \implies Y(s) = \dfrac1{s-1} - \dfrac1s + \dfrac1{4s^2}[/tex]
Taking the inverse transform of both sides, we get
[tex]\boxed{y(t) = e^t - 1 + \dfrac14 t}[/tex]