Solve the System of Equations y=x^2-5x+3 y=x-6

$y = x^{2} - 5 x + 3$ $y = x - 6$

Eliminate the equal sides of each equation and combine.

$x^{2} - 5 x + 3 = x - 6$

Solve $x^{2} - 5 x + 3 = x - 6$ for $x$ .

Move all terms containing $x$ to the left side of the equation.

Subtract $x$ from both sides of the equation.

$x^{2} - 5 x + 3 - x = - 6$

Subtract $x$ from $- 5 x$ .

$x^{2} - 6 x + 3 = - 6$

Add $6$ to both sides of the equation.

$x^{2} - 6 x + 3 + 6 = 0$

Add $3$ and $6$ .

$x^{2} - 6 x + 9 = 0$

Factor using the perfect square rule.

Rewrite $9$ as $3^{2}$ .

$x^{2} - 6 x + 3^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 x = 2 \cdot x \cdot 3$

Rewrite the polynomial.

$x^{2} - 2 \cdot x \cdot 3 + 3^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 3$ .

${(x - 3)}^{2} = 0$

Set the $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Evaluate $y$ when $x = 3$ .

Substitute $3$ for $x$ .

$y = (3) - 6$

Substitute $3$ for $x$ in $y = (3) - 6$ and solve for $y$ .

Remove parentheses.

$y = 3 - 6$

Remove parentheses.

$y = (3) - 6$

Subtract $6$ from $3$ .

$y = - 3$

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(3, - 3)$

The result can be shown in multiple forms.

Point Form:

$(3, - 3)$

Equation Form:

$x = 3, y = - 3$