how do you solve this system? I'm using substitution..
y=3x+4
y=-x^2
First I combined it: -x^2=3x+4
I don't know how to solve it now..
make it a quadratic equation..
x^2+3x+4=0
Now, use the quadratic formula. I think it will have complex roots.
Get all of the terms to the left side by adding -3x and -4 to both sides.
-x^2 - 3x - 4 = 0
divide through by -1 (divide each term by -1)
x^2 + 3x + 4 = 0
factor the trinomial
(x + 3)(x + 1) = 0
see if you can take it from there.
Rick, that factorization is wrong.
To solve this system of equations using substitution, you can substitute the value of y from the first equation into the second equation to eliminate y.
From the first equation, y = 3x + 4. Now substitute this expression for y in the second equation:
-x^2 = 3x + 4
Rearranging the equation, we get:
x^2 + 3x + 4 = 0
At this point, we have a quadratic equation. To solve it, you can either factor it or use the quadratic formula.
Let's try factoring:
The quadratic equation x^2 + 3x + 4 = 0 cannot be factored easily, so we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = 1, b = 3, and c = 4. Plugging these values into the quadratic formula:
x = (-3 ± √(3^2 - 4(1)(4))) / (2(1))
x = (-3 ± √(9 - 16)) / 2
x = (-3 ± √(-7)) / 2
Since we have a negative value under the square root, it means that the quadratic equation has no real solutions. Therefore, this system of equations has no real solutions.
The steps above explain how to solve the system of equations using substitution and how to proceed with solving the resulting equation.