Solving Simultaneous equation

Y=x-2
Y=x^2+3x-5

use substitution

x^2 + 3x - 5 = x-2
x^2 + 2x - 3 = 0
(x+3)(x-1) = 0
x = -3 or x = 1

check with Wolfram
http://www.wolframalpha.com/input/?i=Y%3Dx-2+%2C+Y%3Dx%5E2%2B3x-5

if x = -3, y = -3-2 = -5
if x = 1 , y = 1-2 = -1

x=-3, y=-5 ---> point (-3,-5)
x = 1 , y = -1 --->point (1, -1)

To solve the simultaneous equations:

1. Start by setting the two equations equal to each other since they both equal to "y":
x - 2 = x^2 + 3x - 5

2. Rearrange the equation to bring all the terms to one side to get a quadratic equation:
x^2 + 2x - 3 = 0

3. To solve the quadratic equation, we can use factoring, quadratic formula, or completing the square method. In this case, we will use factoring.

Factor the quadratic equation:
(x + 3)(x - 1) = 0

4. Set each factor equal to zero and solve for x:
x + 3 = 0 or x - 1 = 0

Solving the first equation:
x + 3 = 0
x = -3

Solving the second equation:
x - 1 = 0
x = 1

5. Therefore, the solutions for x are x = -3 and x = 1.

6. Now, substitute the values of x back into either equation to find the corresponding values of y.

For x = -3:
y = x - 2
y = -3 - 2
y = -5

For x = 1:
y = x - 2
y = 1 - 2
y = -1

7. Therefore, the solutions to the simultaneous equations are (-3, -5) and (1, -1).