Solve the System of Equations y=x^2-6x+6 y-x=-4

$y = x^{2} - 6 x + 6$ $y - x = - 4$

Replace all occurrences of $y$ with $x^{2} - 6 x + 6$ in each equation.

Replace all occurrences of $y$ in $y - x = - 4$ with $x^{2} - 6 x + 6$ .

$(x^{2} - 6 x + 6) - x = - 4$

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

Simplify the left side.

Subtract $x$ from $- 6 x$ .

$x^{2} - 7 x + 6 = - 4$

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

$x^{2} - 7 x + 6 = - 4$

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

$x^{2} - 7 x + 6 = - 4$

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

Solve for $x$ in $x^{2} - 7 x + 6 = - 4$ .

Add $4$ to both sides of the equation.

$x^{2} - 7 x + 6 + 4 = 0$

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

Add $6$ and $4$ .

$x^{2} - 7 x + 10 = 0$

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

Factor $x^{2} - 7 x + 10$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $10$ and whose sum is $- 7$ .

$- 5, - 2$

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

Write the factored form using these integers.

$(x - 5) (x - 2) = 0$

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

$(x - 5) (x - 2) = 0$

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

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 5 = 0$

$x - 2 = 0$

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

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

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

Add $5$ to both sides of the equation.

$x = 5$

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

$x = 5$

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

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

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

Add $2$ to both sides of the equation.

$x = 2$

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

$x = 2$

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

The final solution is all the values that make $(x - 5) (x - 2) = 0$ true.

$x = 5, 2$

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

$x = 5, 2$

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

Replace all occurrences of $x$ with $5$ in each equation.

Replace all occurrences of $x$ in $y = x^{2} - 6 x + 6$ with $5$ .

$y = {(5)}^{2} - 6 \cdot 5 + 6$

$x = 5$

Simplify the right side.

Simplify ${(5)}^{2} - 6 \cdot 5 + 6$ .

Simplify each term.

Raise $5$ to the power of $2$ .

$y = 25 - 6 \cdot 5 + 6$

$x = 5$

Multiply $- 6$ by $5$ .

$y = 25 - 30 + 6$

$x = 5$

$y = 25 - 30 + 6$

$x = 5$

Simplify by adding and subtracting.

Subtract $30$ from $25$ .

$y = - 5 + 6$

$x = 5$

Add $- 5$ and $6$ .

$y = 1$

$x = 5$

$y = 1$

$x = 5$

$y = 1$

$x = 5$

$y = 1$

$x = 5$

$y = 1$

$x = 5$

Replace all occurrences of $x$ with $2$ in each equation.

Replace all occurrences of $x$ in $y = x^{2} - 6 x + 6$ with $2$ .

$y = {(2)}^{2} - 6 \cdot 2 + 6$

$x = 2$

Simplify the right side.

Simplify ${(2)}^{2} - 6 \cdot 2 + 6$ .

Simplify each term.

Raise $2$ to the power of $2$ .

$y = 4 - 6 \cdot 2 + 6$

$x = 2$

Multiply $- 6$ by $2$ .

$y = 4 - 12 + 6$

$x = 2$

$y = 4 - 12 + 6$

$x = 2$

Simplify by adding and subtracting.

Subtract $12$ from $4$ .

$y = - 8 + 6$

$x = 2$

Add $- 8$ and $6$ .

$y = - 2$

$x = 2$

$y = - 2$

$x = 2$

$y = - 2$

$x = 2$

$y = - 2$

$x = 2$

$y = - 2$

$x = 2$

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

$(5, 1)$

$(2, - 2)$

The result can be shown in multiple forms.

Point Form:

$(5, 1), (2, - 2)$

Equation Form:

$x = 5, y = 1$

$x = 2, y = - 2$

Solve the System of Equations y=x^2-6x+6 y-x=-4

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