To solve this system of equations, we can substitute the expression for Y from the second equation into the first equation.
Substituting Y = x + 2 into the first equation, we get:
x + 2 = x^2 + 5x - 3
Rearranging the equation to set it equal to zero:
x^2 + 4x - 5 = 0
From there, we can factor the quadratic equation:
(x + 5)(x - 1) = 0
Setting each factor equal to zero:
x + 5 = 0 or x - 1 = 0
Solving for x, we have:
x = -5 or x = 1
Substituting these values back into either of the original equations, we can solve for Y.
When x = -5:
Y = (-5)^2 + 5(-5) - 3 = 25 - 25 - 3 = -3
When x = 1:
Y = (1)^2 + 5(1) - 3 = 1 + 5 - 3 = 3
Thus, the solution to the system of equations is x = -5, Y = -3 or x = 1, Y = 3.
Solve the system of equations.
Y=x^(2)+5x-3
Y-x=2
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