Answer:x = (-7 ± √(89 + 4y)) / 2
x = (-2 ± √(12y - 68)) / 6
Step-by-step explanation:To solve the system of equations algebraically for x, let's start by rearranging both equations to isolate the variable x.
1. Given equation: x^2 + 7x = 5 + y
- Rearrange the equation: x^2 + 7x - (5 + y) = 0
2. Given equation: 2x - 3x^2 = y - 9
- Rearrange the equation: 3x^2 + 2x - (y - 9) = 0
Now, we have two quadratic equations in the form ax^2 + bx + c = 0.
To solve these equations, we can use either factoring, completing the square, or the quadratic formula. However, since these equations are not easily factorable, we will use the quadratic formula.
The quadratic formula is given as: x = (-b ± √(b^2 - 4ac)) / (2a)
For the first equation, x^2 + 7x - (5 + y) = 0, the values of a, b, and c are:
a = 1
b = 7
c = -(5 + y)
Applying the quadratic formula, we have:
x = (-7 ± √(7^2 - 4(1)(-(5 + y)))) / (2(1))
x = (-7 ± √(49 + 20 + 4y + 20)) / 2
x = (-7 ± √(89 + 4y)) / 2
For the second equation, 3x^2 + 2x - (y - 9) = 0, the values of a, b, and c are:
a = 3
b = 2
c = -(y - 9)
Applying the quadratic formula, we have:
x = (-2 ± √(2^2 - 4(3)(-(y - 9)))) / (2(3))
x = (-2 ± √(4 + 12y - 72)) / 6
x = (-2 ± √(12y - 68)) / 6
So, the solutions for x in the system of equations are:
x = (-7 ± √(89 + 4y)) / 2
x = (-2 ± √(12y - 68)) / 6