What are the solutions of the system?
y = x2 + 3x – 4
y = 2x + 2
Don't understand
Eq1: Y = x^2+3x-4
Eq2: Y = 2x+2
In Eq1, replace Y with 2x+2 and solve for X:
2x+2 = x^2+3x-4
x^2+3x-2x = 2+4
x^2+x = 6
x^2+x-6 = 0
C = -6 = -2*3.
(x-2)(x+3) = 0
x-2 = 0, X = 2.
x+3 = 0, X = -3.
Solution: X = -3, and 2.
To find the solutions of this system of equations, we need to solve for the values of x and y that satisfy both equations simultaneously. We can do this by using the method of substitution or elimination.
One way to solve this system is by using the method of substitution.
First, rearrange the second equation to isolate one variable in terms of the other variable. In this case, solve the second equation for y:
y = 2x + 2
Next, substitute this expression for y into the first equation:
x^2 + 3x - 4 = 2x + 2
Now, we have a quadratic equation in terms of x. Rearrange the equation to one side:
x^2 + x - 6 = 0
We can solve this quadratic equation by factoring or using the quadratic formula. Factoring gives us:
(x + 3)(x - 2) = 0
Setting each factor equal to zero:
x + 3 = 0 or x - 2 = 0
Solving for x gives us two possible values:
x = -3 or x = 2
Now, substitute these values of x back into either of the original equations to find the corresponding y values:
For x = -3:
y = 2(-3) + 2
y = -6 + 2
y = -4
For x = 2:
y = 2(2) + 2
y = 4 + 2
y = 6
Therefore, the solutions of the system are (-3, -4) and (2, 6).