To solve the system of equations, we need to equate the two equations and solve for x.
Setting the two equations equal to each other, we have:
x^2 + 12x + 13 = 3x - 5
Rearranging the equation, we get:
x^2 + 12x - 3x + 13 + 5 = 0
Combining like terms, we have:
x^2 + 9x + 18 = 0
To solve this quadratic equation, we can factor it:
(x + 3)(x + 6) = 0
Setting each factor equal to zero, we get:
x + 3 = 0 or x + 6 = 0
Solving for x, we have:
x = -3 or x = -6
Now that we have the values of x, we can substitute them back into one of the original equations to find the corresponding values of y.
For x = -3:
y = (-3)^2 + 12(-3) + 13
y = 9 - 36 + 13
y = -14
Therefore, one solution to the system of equations is x = -3 and y = -14.
For x = -6:
y = (-6)^2 + 12(-6) + 13
y = 36 - 72 + 13
y = -23
Therefore, another solution to the system of equations is x = -6 and y = -23.
The system of linear and quadratic equations is solved, and the solutions are x = -3, y = -14 and x = -6, y = -23.
solve the system of linear and quadratic equations.
y=x^2+12x+13
y=3x-5
1 answer
1 answer