Question

What are the solutions to the system of equations?
x = x^2 - 4x +3
y = -x +3

Answer 1

For this case we have the following system of equations:

`y = x ^ 2-4x + 3\\y = -x + 3`

Matching we have:

`x ^ 2-4x + 3 = -x + 3\\x ^ 2-4x + x + 3-3 = 0\\x ^ 2-3x = 0`

We solve by means of

`x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}`

Where:

`a = 1\\b = -3\\c = 0`

Substituting:

`x = \frac {- (- 3) \pm \sqrt {(- 3) ^ 2-4 (1) (0)}} {2 (1)}\\x = \frac {3 \pm \sqrt {9}} {2}\\x = \frac {3 \pm3} {2}`

Finally, the roots are:

`x_ {1} = \frac {3-3} {2} = 0\\x_ {2} = \frac {3 + 3} {2} = \frac {6} {2} = 3`

Answer:

`x_ {1} = 0\\x_ {2} = 3`