solve the system of linear and quadratic equations.

Question

y=x^2+12x+13
y=3x-5

Answer 1

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.