What are the solution of the system?
y=x^2+3x-4
y=3x+2
A (-3,6) and (2,-4)
B(-3,-4) and (2,6)
C (-3,-4) and (-2,-2)
D no solution
I think it is D
x^2+3x-4 = 3x+2
x^2 = 6
x = ±√6
If x = √6 , y = 3√6 + 2
if x = -√6 , y = -3√6 + 2
none of the choices given are correct, do you have a typo?
http://www.wolframalpha.com/input/?i=solve++y%3Dx%5E2%2B3x-4+,+y%3D3x%2B2
3x+2 = x^2+3x-4
x^2 - 6 = 0
x = +/- sqrt 6
if +sqrt 6
y = 3sqrt 6 + 2 = 6+3sqrt 6 -4 check :)
there are solutions but none of them are on your list
To find the solutions of the system, we need to solve the two equations simultaneously.
First, let's rewrite the equations in the standard form, which is y = mx + b, where m is the slope and b is the y-intercept.
Equation 1: y = x^2 + 3x - 4
Equation 2: y = 3x + 2
Now, set the two equations equal to each other:
x^2 + 3x - 4 = 3x + 2
Next, simplify the equation:
x^2 + 3x - 4 - 3x - 2 = 0
x^2 - 6 = 0
To solve this quadratic equation, we can factor or use the quadratic formula. However, in this case, the equation can be factored easily:
(x - 2)(x + 2) = 0
This means that either x - 2 = 0 or x + 2 = 0. Solve each equation separately:
Case 1: x - 2 = 0
x = 2
Case 2: x + 2 = 0
x = -2
Now, substitute these values back into either of the original equations to find the corresponding y-values.
Using Equation 2: y = 3x + 2
For x = 2:
y = 3(2) + 2
y = 6 + 2
y = 8
For x = -2:
y = 3(-2) + 2
y = -6 + 2
y = -4
Therefore, the solutions of the system are (-2, -4) and (2, 8), which is not one of the given answer choices. Since none of the answer choices match the solutions we found, we can conclude that the correct answer would be D: no solution.