Solve the following system of equations algebraically:y, equals, x, squared, minus, 2, x, minus, 43y=x 2 −2x−43y, equals, 2, x, plus, 2y=2x+2
Question
Solve the following system of equations algebraically:y, equals, x, squared, minus, 2, x, minus, 43y=x 2 −2x−43y, equals, 2, x, plus, 2y=2x+2
Solution
The system of equations is:
- y = x^2 - 2x - 43
- y = 2x + 2
To solve this system algebraically, we can set the two equations equal to each other because they both equal y:
x^2 - 2x - 43 = 2x + 2
Next, we can simplify this equation by subtracting 2x and 2 from both sides:
x^2 - 2x - 2x - 43 - 2 = 0 x^2 - 4x - 45 = 0
This is a quadratic equation in the form ax^2 + bx + c = 0. We can solve for x using the quadratic formula x = [-b ± sqrt(b^2 - 4ac)] / (2a):
x = [4 ± sqrt((-4)^2 - 41(-45))] / (2*1) x = [4 ± sqrt(16 + 180)] / 2 x = [4 ± sqrt(196)] / 2 x = [4 ± 14] / 2 x = 9 or x = -5
Substitute x = 9 and x = -5 into the second equation y = 2x + 2 to solve for y:
When x = 9, y = 29 + 2 = 20 When x = -5, y = 2(-5) + 2 = -8
So, the solutions to the system of equations are (9, 20) and (-5, -8).
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