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Let X(s) and Y(s) denote the Laplace transforms of x(t) and y(t), respectively. Then taking the transform of both sides of both equations gives

LT[dx/dt + 5x + dy/dt] = LT[1]

==>   s X(s) - x (0) + 5 X(s) + s Y(s) - y (0) = 1/s

==>   s X(s) + 5 X(s) + s Y(s) = 1/s

==>   (s + 5) X(s) + s Y(s) = 1/s

LT[dx/dt - x + dy/dt - y] = LT[exp(t )]

==>   s X(s) - x (0) - X(s) + s Y(s) - y (0) - Y(s) = 1/(s - 1)

==>   s X(s) - X(s) + s Y(s) - Y(s) = 1/(s - 1)

==>   (s - 1) X(s) + (s - 1) Y(s) = 1/(s - 1)

Solve for X(s) and Y(s). Using elimination, you would get

X(s) = (1 - 2s) / (5s (s - 1)²)

Y(s) = (7s - 1) / (5s (s - 1)²)

Now take the inverse transforms of each. Start by getting the partial fraction decompositions:

(1 - 2s) / (5s (s - 1)²) = 1/5 (a/s + b/(s - 1) + c/(s - 1)²)

-2s + 1 = a (s - 1)² + bs (s - 1) + cs

-2s + 1 = (a + b) s ² + (-2a - b + c) s + a

==>   a + b = 0, -2a - b + c = -10, a = 5

==>   a = 1, b = -1, c = -1

==>   X(s) = 1/5 (1/s - 1/(s - 1) - 1/(s - 1)²)

Similarly, you would find

Y(s) = -1/5 (1/s - 1/(s - 1) - 6/(s - 1)²)

Now for the inverse transforms:

LT⁻¹ [1/s] = 1

LT⁻¹ [1/(s - 1)] = exp(t )

LT⁻¹ [1/(s - 1)²] = t exp(t )

Putting everything together, we have

LT⁻¹ [X(s)] = x(t) = 1/5 - 1/5 exp(t ) - 1/5 t exp(t )

and

LT⁻¹ [Y(s)] = y(t) = -1/5 + 1/5 exp(t ) + 6/5 t exp(t )