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What is the solution of the linear-quadratic system of equations?
{y=x2+5x−3
y−x=2
To find the solution of the linear-quadratic system of equations, we can substitute the value of y from the second equation into the first equation.
From the second equation, we have y - x = 2.
Rearranging this equation, we get y = x + 2.
Now, we substitute this value of y into the first equation:
x^2 + 5x - 3 = x + 2.
Rearranging the equation, we get:
x^2 + 4x - 5 = 0.
This is a quadratic equation that can be factored:
(x + 5)(x - 1) = 0.
Setting each factor equal to zero, we have two possible solutions:
x + 5 = 0 --> x = -5
x - 1 = 0 --> x = 1
Now, we substitute these values back into the second equation to find the corresponding values of y:
For x = -5:
y - (-5) = 2
y + 5 = 2
y = -3
For x = 1:
y - 1 = 2
y = 3
Therefore, the solution of the linear-quadratic system of equations is:
(x, y) = {(-5, -3), (1, 3)}