Solve the following system of equations algebraically:y, equals, x, squared, plus, 4, x, plus, 3y=x 2 +4x+3y, equals, minus, x, plus, 9y=−x+9

The system of equations is:

1. y = x^2 + 4x + 3
2. y = -x + 9

To solve this system algebraically, we can set the two equations equal to each other because they both equal y:

x^2 + 4x + 3 = -x + 9

Next, we can simplify this equation by adding x to both sides and subtracting 9 from both sides:

x^2 + 5x - 6 = 0

This is a quadratic equation, which we can solve by factoring:

(x + 6)(x - 1) = 0

Setting each factor equal to zero gives the solutions x = -6 and x = 1.

We can then substitute these values back into the original equations to find the corresponding y values. For x = -6:

y = (-6)^2 + 4*(-6) + 3 = 36 - 24 + 3 = 15

And for x = 1:

y = 1^2 + 4*1 + 3 = 1 + 4 + 3 = 8

So the solutions to the system of equations are (-6, 15) and (1, 8).