The solution to the system of simultaneous equations is; (-1, 4) and (3, 0)
What is the solution to the quadratic equation?
We are given two equations as;
y = x² - x - 6 -----(1)
y = x - 3 ----(2)
We will use the algebraic method to solve it as;
The general format for a quadratic equation is;
y = ax² + bx + c
The quadratic formula to solve the quadratic equation is;
x = [-b ± √(b² - 4ac)]/(2a)
Put equation 2 into equation 1 to get;
x - 3 = x² - x - 6
x² - x - x - 6 + 3 = 0
x² - 2x - 3 = 0
Factorizing we have;
x² + x - 3x - 3 = 0
x(x + 1) - 3(x + 1) = 0
Thus;
x - 3 = 0 or x + 1 = 0
x = -1 or 3
Read more about Quadratic Equation Solution at; https://brainly.com/question/1214333
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