Solve the pair of simultaneous equation

Y=xsquared-x+3
Y=6-3x

y=x^2-x+3

y=6-3x
so, set them equal
x^2-x+3 = 6-3x
x^2+2x-3 = 0
(x+3)(x-1) = 0
x = 1 or -3

you probably want to wait for one of the admins to check it, but I think the answer may be 1. if x^2 + 2x=3, the only one that makes it true is 1.

y = x^2 - x + 3

3 x = 6-y or x = 2 - y/3
so
x^2 = 4 -4y/3 + y^2/9
then y = 4 - 4y/3 + y^2/9 -2 +y/3 + 3
y^2/9 - 2 y + 7 = 0
y^2 - 18 y + 63 = 0
(y-21)(y+3) = 0

I get 21 and -3

but matty is right, 1 works

Thanks much couldn't have done it without ur help.. no fully understand but thanks..

To solve the pair of simultaneous equations:

1. Set the two equations equal to each other:
xsquared - x + 3 = 6 - 3x

2. Rearrange the equation to get all the variables on one side:
xsquared - x + 3x - 6 = 0

3. Combine like terms:
xsquared + 2x - 6 = 0

4. Now, we need to solve this quadratic equation using factoring, completing the square, or the quadratic formula. In this case, let's use factoring.

a) Write the equation in the form: (x - )(x + ) = 0, where we need to find the factors of the constant term (-6) that add up to the coefficient of x (2).

Factors of -6: -1, 1, -2, 2, -3, 3, -6, 6
-1 and 6 are the factors that add up to 2.

b) Rewrite the equation as:
(x - 1)(x + 6) = 0

5. Set each factor equal to zero and solve for x:
x - 1 = 0 or x + 6 = 0

Solve for x in each equation:
x = 1 or x = -6

6. Now that we have the values of x, substitute them back into either of the original equations to find the corresponding values of y.

For x = 1:
Y = (1^2) - 1 + 3
= 1 - 1 + 3
= 3

For x = -6:
Y = (-6^2) - (-6) + 3
= 36 + 6 + 3
= 45

Therefore, the solutions to the simultaneous equations are:
x = 1, y = 3
x = -6, y = 45