What is the solution of the linear-quadratic system of equations?
y=x2+5x−3
{
y−x=2
To solve this system of equations, we first need to substitute the expression for y from the first equation into the second equation.
From the first equation, we have y = x^2 + 5x - 3.
Substitute this expression into the second equation:
x^2 + 5x - 3 - x = 2
x^2 + 4x - 3 = 2
x^2 + 4x - 5 = 0
Now, we need to solve this quadratic equation for x. We can use the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
In this case, a = 1, b = 4, and c = -5.
x = (-4 ± sqrt(4^2 - 4*1*(-5))) / 2*1
x = (-4 ± sqrt(16 + 20)) / 2
x = (-4 ± sqrt(36)) / 2
x = (-4 ± 6) / 2
This gives us two possible values of x:
x = (2/2) = 1
x = (-10/2) = -5
Now that we have the values of x, we can substitute them back into the first equation to find the corresponding values of y:
For x = 1:
y = 1^2 + 5(1) - 3 = 1 + 5 - 3 = 3
For x = -5:
y = (-5)^2 + 5(-5) - 3 = 25 - 25 - 3 = -3
Therefore, the solutions of the linear-quadratic system of equations are:
x = 1, y = 3
x = -5, y = -3