Simultaneous equations...
y = x^2 - x - 1
y = 2 - 3x
Many thanks.
Since both equations = y then they must be equal to each other.
x^2 - x-1 = 2 - 3x
To solve, you need all terms on one side and zero on the other.
x^2+2x -3=0
Factor
(x+3)(x-1) = 0
so x = -3 and x = 1
now to find y.. substitute -3 for x in each equation and then 1 for x in each equation.
This is different from straight lines which only cross in one place. You have a parabola and a straight line so they will cross in two places.
How did you get +2x would it not be -2x?
x^2 - x - 1 = 2 - 3x
Add x to each side...
x^2 - 1 = 2 - 2x
Then add 1 to each side and that makes the equation...
x^2 = 3 -2x
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you can do it that way, but how has that helped?
How do you solve x^2 = 3-2x ?
To solve this system of simultaneous equations, we can use the method of substitution or elimination. Let's start by using substitution:
1. Start with the first equation: y = x^2 - x - 1.
2. Substitute this expression for y in the second equation: x^2 - x - 1 = 2 - 3x.
3. Rearrange the equation to bring all terms to one side: x^2 - x + 3x - 1 - 2 = 0.
4. Simplify the equation: x^2 + 2x - 3 = 0.
5. Factorize the quadratic equation: (x + 3)(x - 1) = 0.
6. Set each factor equal to zero: x + 3 = 0 or x - 1 = 0.
7. Solve for x: x = -3 or x = 1.
Now that we have the values for x, we can substitute them back into either equation to find the corresponding values for y:
If x = -3, y = 2 - 3(-3) = 2 + 9 = 11.
If x = 1, y = 2 - 3(1) = 2 - 3 = -1.
Therefore, the solution to the simultaneous equations is x = -3, y = 11 and x = 1, y = -1.