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What is the solution of the linear-quadratic system of equations?
{y=x2+5x−3y−x=2
To solve this system of equations, we need to find the values of x and y that satisfy both equations simultaneously.
1. Substitute the value of y from the second equation into the first equation:
x^2 + 5x - 3(y - x) = 2
x^2 + 5x - 3y + 3x = 2
x^2 + 8x - 3y = 2
2. Substitute the value of y from the first equation into the second equation:
y = x^2 + 5x - 3
x^2 + 5x - 3 - x = 2
x^2 + 4x - 5 = 0
3. This equation is a quadratic equation in standard form. We can use the quadratic formula to solve for x:
x = (-b ± √(b^2 - 4ac)) / 2a
x = (-4 ± √(4^2 - 4*1*(-5))) / 2*1
x = (-4 ± √(16 + 20)) / 2
x = (-4 ± √36) / 2
x = (-4 ± 6) / 2
4. Therefore, the possible solutions for x are:
x = (-4 + 6) / 2 = 1
x = (-4 - 6) / 2 = -5
5. Now, substitute these values back into the first equation to find the corresponding values of y:
For x = 1:
y = 1^2 + 5*1 - 3
y = 1 + 5 - 3
y = 3
For x = -5:
y = (-5)^2 + 5*(-5) - 3
y = 25 - 25 - 3
y = -3
6. Therefore, the solutions to the linear-quadratic system of equations are (1, 3) and (-5, -3).