I have to solve a few systems through elimination, but I'm still not quite sure how to do it..
The first is y=x^2 and y=x+2
Can someone show me how exactly it is worked out so I can try the second one myself?
x^2 = x + 2
x^2 - x - 2 = 0
(x - 2)(x + 1) = 0
x - 2 = 0 ... x = 2
x + 1 = 0 ... x = -1
the line intersects the parabola in two places
plug the x-values into the equations to find the corresponding y-values
I understand how to do that, thats substitution. How can I solve it with elimination?
if you subtract the equations (to eliminate y), you still end up with
... 0 = x^2 - x - 2
... but you have used elimination
even with elimination, you have to substitute back at some point to find all the variables
Sorry :) I must've gotten something confused. I think I get it now. Thank you!
Ok.. I understand how to eliminate y and everything, then you factor it as normal. The solutions to that equation were (-1, 1) and (2,4).
I'm working on another one and used the same steps.
y=48x+20
-(y=-x^2+200x+20)
and i ended up with :
0=x^2-152x
This can't be factored, does that mean it has no solution?
*facepalm* That tricked me. I think it's (152, 7316) & (0,20)? I guess ANY number times 0 is a factor of 0... So I did -152 x 0=0 and -152+0 = -152.
That gave me (x-152)(x+0)
Sure, I can help you with that! To solve a system of equations through elimination, you need to manipulate the equations in a way that one of the variables cancels out when you add or subtract the equations.
Let's start with the given system of equations:
1) y = x^2
2) y = x + 2
To eliminate one of the variables, you need to make the coefficients of either x or y the same in both equations. In this case, it's easier to eliminate y.
Since both equations have y by itself on one side, you can set them equal to each other:
x^2 = x + 2
Now you have a quadratic equation. You can solve it by bringing all terms to one side and solving for x:
x^2 - x - 2 = 0
To factor or use the quadratic formula, we notice that the equation can be factored as:
(x - 2)(x + 1) = 0
Setting each factor equal to zero gives us two possible solutions for x:
x - 2 = 0 or x + 1 = 0
Solving these equations, we find:
x = 2 or x = -1
Now that we have the values for x, we can substitute them back into one of the original equations to find the corresponding values of y.
If we substitute x = 2 into equation 2), we get:
y = 2 + 2
y = 4
So the first solution is x = 2, y = 4.
If we substitute x = -1 into equation 2), we get:
y = -1 + 2
y = 1
So the second solution is x = -1, y = 1.
Therefore, the system of equations y = x^2 and y = x + 2 has two solutions: (2,4) and (-1,1).
Now, feel free to try the second system of equations on your own! If you have any further questions, please let me know.