Gosh, I dislike the "Laplace transform". I have never seen a problem that couldn't be done more easily using direct methods rather than "Laplace transform"! I suspect the "Laplace transform" just makes engineers feel better since it only involves "plugging and cranking" rather than thinking!
In this case, given x'= 2x- y and y'= 5x- 2y, differentiating, y''= 5x'- 2y'= 5(2x- y)- 2y= 10x- 7y. From y'= 5x- 2y, 5x= y'+ 2y so 10x= 2y'+t 4y. y''= 2y'+ 4y- 7y= 2y'- 3y. y''- 2y'- 3y= 0 has characteristic equation \(\displaystyle r^2- 2r- 3= (r- 3)(r+ 1)= 0\). r= 3 and -1 so the general solution for y is \(\displaystyle y(t)= C_1e^{3t}+ C_2e^{-t}\).
x(t) can then be solved from \(\displaystyle 5x= y'+ 2y\): \(\displaystyle y(t)= C_1e^{3t}+ C_2e^{-t}\) so \(\displaystyle y'(t)= 3C_1r^{3t}- C_2e^{-t}\). \(\displaystyle 5x= 3C_1e^{3t}- C_2e^{-t}+ 2C_1e^{3t}+ 2C_2e^{-t}= 5C_1e^{3t}+ C_2e^{-t}\) so \(\displaystyle x(t)= C_1e^{3t}+ \frac{C_2}{5}e^{-t}\).
Now, \(\displaystyle x(0)= C_1+ \frac{C_2}{5}= 1\) and \(\displaystyle y(0)= C_1+ C_2= 1\). Subtracting the first equation from the second, \(\displaystyle \frac{4C_2}{5}= 0\) so \(\displaystyle C_2= 0\). Then \(\displaystyle C_1= 1\).
So the solution to this system of equations is
\(\displaystyle x(t)= e^{3t}\) and \(\displaystyle y(t)= e^{3t}\).