Solve the following system of equations algebraically: y, equals, x, squared, minus, 7, x, minus, 7 y= x 2 −7x−7 y, equals, minus, 3, x, minus, 2 y= −3x−2

Answer 1

To solve this system of equations algebraically, we will set the two equations equal to each other and solve for x.

x^2 - 7x - 7 = -3x - 2

Now, we will simplify and rearrange the equation:

x^2 - 7x + 3x - 7 + 2 = 0
x^2 - 4x - 5 = 0

Next, we will factor the quadratic equation:

(x - 5)(x + 1) = 0

Setting each factor equal to zero:

x - 5 = 0 or x + 1 = 0

x = 5 or x = -1

Now that we have the values of x, we can substitute them back into one of the original equations to solve for y. Let's choose y = x^2 - 7x - 7:

1. When x = 5:
y = (5)^2 - 7(5) - 7
y = 25 - 35 - 7
y = -17

Therefore, the solution for x = 5 and y = -17

2. When x = -1:
y = (-1)^2 - 7(-1) - 7
y = 1 + 7 - 7
y = 1

Therefore, the solution for x = -1 and y = 1

Therefore, the solutions for this system of equations are x = 5, y = -17 and x = -1, y = 1.