Question

Consider the following system of equations:


-x + y = 5

y = (x-3)2 + 2

Answer 1

To find the solution to this system, we can substitute the second equation into the first equation and solve for x.

-x + (x-3)2 + 2 = 5

Simplifying the equation, we obtain:

- x + x^2 - 6x + 9 + 2 = 5

Combining like terms, we have:

x^2 - 7x + 6 = 0

Now, we need to solve this quadratic equation. Factoring the equation, we get:

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

Setting each factor equal to zero, we have:

x-6 = 0 or x-1 = 0

Solving for x in each case, we find:

x = 6 or x = 1

Now, we substitute these values of x back into the second equation to find the corresponding values of y.

For x = 6:

y = (6-3)2 + 2

y = 3^2 + 2

y = 9 + 2

y = 11

So, one solution to the system is x = 6, y = 11.

For x = 1:

y = (1-3)2 + 2

y = (-2)^2 + 2

y = 4 + 2

y = 6

So, the other solution to the system is x = 1, y = 6.

Therefore, the system of equations has two solutions: (6, 11) and (1, 6).