y=-3over 2x -2
2x-y=-5
To solve this system of equations, we can use the substitution method.
1. Start with the equation y = -3/(2x - 2) and substitute it into the second equation:
2x - (-3/(2x - 2)) = -5
2x + 3/(2x - 2) = -5
2. Multiply through by (2x - 2) to get rid of the denominator:
2x(2x - 2) + 3 = -5(2x - 2)
4x^2 - 4x + 3 = -10x + 10
4x^2 - 6x + 3 = 0
3. Solve for x using the quadratic formula:
x = [6 ± √(6^2 - 4(4)(3))] / (2(4))
x = [6 ± √(36 - 48)] / 8
x = [6 ± √(-12)] / 8
x = [6 ± 2√3 i] / 8
x = (3 ± √3 i) / 4
4. Now that we have the value of x, we can substitute it back into one of the original equations to solve for y:
Plugging x = (3 ± √3 i) / 4 into y = -3/(2x - 2):
x = (3 + √3 i) / 4:
y = -3 / [2(3 + √3 i) - 2]
y = -3 / [6 + 2√3 i - 2]
y = -3 / [4 + 2√3 i]
y = -3 / 4 - 3√3 i
y = -3 / 4 - 3√3 i
x = (3 - √3 i) / 4:
y = -3 / [2(3 - √3 i) - 2]
y = -3 / [6 - 2√3 i - 2]
y = -3 / [4 - 2√3 i]
y = -3 / 4 + 3√3 i
y = -3 / 4 + 3√3 i
Therefore, the solutions are (-3 + 3√3 i) / 4 and (-3 - 3√3 i) / 4.