To solve the system of equations using the substitution method, we'll set the two equations equal to each other because they both equal y:
x^2 - 5x - 5 = -2x^2 + 6x - 1
Now we'll combine like terms by bringing all terms to one side of the equation:
x^2 + 2x^2 - 5x - 6x - 5 + 1 = 0
3x^2 - 11x - 4 = 0
We have a quadratic equation, which we will attempt to solve for x. This equation can be factored, and if it cannot be factored easily, you could use the quadratic formula.
3x^2 - 12x + x - 4 = 0
Group the terms:
(3x^2 - 12x) + (x - 4) = 0
Factor out the common factors:
3x(x - 4) + 1(x - 4) = 0
Now factor by grouping:
(3x + 1)(x - 4) = 0
Set each factor equal to zero to solve for x:
3x + 1 = 0 or x - 4 = 0
x = -1/3 or x = 4
Now we can use these solutions for x to solve for y using either of the two original equations. Let's use the first equation:
For x = -1/3:
y = (-1/3)^2 - 5(-1/3) - 5
y = 1/9 + 5/3 - 5
y = 1/9 + 15/9 - 45/9
y = 16/9 - 45/9
y = -29/9
For x = 4:
y = (4)^2 - 5(4) - 5
y = 16 - 20 - 5
y = -9
So the solutions to the system of equations are:
(x, y) = (-1/3, -29/9) and (x, y) = (4, -9)
Solve the system of equations by substitution.
y=x^2-5x-5
y=-2x^2+6x-1
1 answer