Find solutions to the system
y=x^2+3x-4
y=2x+2
A (-3,6)and(2,-4)
B (-3,-4) and (2,6)
C (-3,-4) and (-2,-2)
D no solutions
To find the solutions to the system of equations, we need to find the values of x and y that satisfy both equations.
Let's assume the system is:
y = x^2 + 3x - 4 ...(Equation 1)
y = 2x + 2 ...(Equation 2)
To find the solutions, we'll substitute Equation 2 into Equation 1:
2x + 2 = x^2 + 3x - 4
Rearranging the equation to the standard quadratic form, we have:
x^2 + x - 6 = 0
Now we'll solve this quadratic equation by factoring or using the quadratic formula. The factored form is:
(x + 3)(x - 2) = 0
Setting each factor equal to zero, we have:
x + 3 = 0 --> x = -3
x - 2 = 0 --> x = 2
Now, we'll substitute these values of x back into one of the original equations to find the corresponding y values.
For x = -3:
y = (-3)^2 + 3(-3) - 4
y = 9 - 9 - 4
y = -4
For x = 2:
y = (2)^2 + 3(2) - 4
y = 4 + 6 - 4
y = 6
Therefore, the solutions to the system of equations are:
A: (-3, -4) and (2, 6)
So, the correct answer is B (-3,-4) and (2,6).
To find the solutions to the system of equations, we need to find the values of x and y that satisfy both equations simultaneously.
Let's start by setting the two equations equal to each other:
x^2 + 3x - 4 = 2x + 2
Next, we'll simplify the equation:
x^2 + 3x - 4 - 2x - 2 = 0
Combining like terms gives us:
x^2 + x - 6 = 0
Now we can solve this quadratic equation. There are multiple methods for solving quadratics, such as factoring, completing the square, or using the quadratic formula. In this case, let's use factoring.
To factor the quadratic equation x^2 + x - 6 = 0, we need to find two numbers that multiply to -6 and add up to 1 (coefficient of x).
The numbers that satisfy this condition are 3 and -2. So we can rewrite the equation as:
(x + 3)(x - 2) = 0
Now we can apply the zero-product property, which states that if the product of two factors is equal to zero, then at least one of the factors must be zero. Therefore, we can set each factor equal to zero and solve for x:
x + 3 = 0 or x - 2 = 0
Solving these equations gives us:
x = -3 or x = 2
Now that we have the x-values, we can substitute them back into either equation to find the corresponding y-values.
Substituting x = -3 into the first equation:
y = (-3)^2 + 3(-3) - 4
y = 9 - 9 - 4
y = -4
So, one solution is (-3, -4).
Substituting x = 2 into the first equation:
y = (2)^2 + 3(2) - 4
y = 4 + 6 - 4
y = 6
So, the other solution is (2, 6).
Therefore, the correct answer is option B: (-3, -4) and (2, 6).
just sub in2x+2 for y
x^2+3x-4 = 2x+2
x^2+x-6 = 0
(x+3)(x-2) = 0
...
or, without solving, you could just check each of the choices... duh