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To solve the system, we can substitute the value of y from the second equation into the first equation.
From the second equation, we have y = x + 2.
Substituting this into the first equation: x + 2 = x^2 + 5x - 3.
Rearranging the equation: x^2 + 4x - 5 = 0.
Factorizing the equation: (x + 5)(x - 1) = 0.
Setting each factor to zero, we get two possible solutions:
x + 5 = 0, which gives x = -5.
x - 1 = 0, which gives x = 1.
Now we can substitute these values of x back into one of the original equations to find the corresponding values of y.
For x = -5: y = (-5)^2 + 5(-5) - 3 = 25 - 25 - 3 = -3.
For x = 1: y = (1)^2 + 5(1) - 3 = 1 + 5 - 3 = 3.
Therefore, the solution to the linear-quadratic system of equations is x = -5, y = -3 and x = 1, y = 3.