solve the following system of equations algebraically or graphically for x and y: y=xsquare-6x+6
y-x=-4
For an algebraicc solution, show your work here.
y=x^2-6x+6
y-x= -4
Let's substitute x-4 for y in the first equation.
x-4 = x^2-6x+6
x^2 -7x + 10 = 0
(x-5)(x-2) = 0
x = 5 or 2
y = 1 (where x=5) or -2 (where x=2)
The solutions to the system of equations are (x=5, y=1) and (x=2, y=-2).
To solve the system of equations algebraically, we start by rearranging the first equation to isolate y:
y = x^2 - 6x + 6
Then, we substitute this expression for y in the second equation:
y - x = -4
(x^2 - 6x + 6) - x = -4
Next, we simplify the equation:
x^2 - 7x + 6 - x = -4
x^2 - 8x + 10 = 0
Now, we can solve this quadratic equation. We have two options:
1) Factoring:
To factorize the quadratic equation, we need to find two numbers that multiply to give us 10 and add up to -8. The two numbers are -5 and -2. Thus, we have:
(x - 5)(x - 2) = 0
Setting each factor to zero, we get:
x - 5 = 0, x - 2 = 0
Solving these equations, we find:
x = 5, x = 2
2) Quadratic Formula:
Alternatively, we can use the quadratic formula to solve for x. The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = -8, and c = 10. Plugging these values into the quadratic formula, we get:
x = (-(-8) ± √((-8)^2 - 4(1)(10))) / (2(1))
x = (8 ± √(64 - 40)) / 2
x = (8 ± √24) / 2
x = (8 ± 2√6) / 2
x = 4 ± √6
So, we have two possible values for x: x = 4 + √6 and x = 4 - √6.
Once we have found the values of x, we can substitute them back into one of the original equations to find the corresponding values of y.
For x = 5:
y = (5)^2 - 6(5) + 6
y = 25 - 30 + 6
y = 1
For x = 2:
y = (2)^2 - 6(2) + 6
y = 4 - 12 + 6
y = -2
Therefore, the solution to the system of equations is (x, y) = {(5, 1), (2, -2)}.