Solve the System of Equations x+y=3 y=x^2-6x+9

$x + y = 3$ $y = x^{2} - 6 x + 9$

Step 1

Replace all occurrences of $y$ with $x^{2} - 6 x + 9$ in each equation.

Replace all occurrences of $y$ in $x + y = 3$ with $x^{2} - 6 x + 9$ .

$x + x^{2} - 6 x + 9 = 3$

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

Simplify the left side.

Simplify $x + x^{2} - 6 x + 9$ .

Remove parentheses.

$x + x^{2} - 6 x + 9 = 3$

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

Subtract $6 x$ from $x$ .

$x^{2} - 5 x + 9 = 3$

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

$x^{2} - 5 x + 9 = 3$

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

$x^{2} - 5 x + 9 = 3$

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

$x^{2} - 5 x + 9 = 3$

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

Step 2

Solve for $x$ in $x^{2} - 5 x + 9 = 3$ .

Subtract $3$ from both sides of the equation.

$x^{2} - 5 x + 9 - 3 = 0$

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

Subtract $3$ from $9$ .

$x^{2} - 5 x + 6 = 0$

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

Factor $x^{2} - 5 x + 6$ 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 $6$ and whose sum is $- 5$ .

$- 3, - 2$

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

Write the factored form using these integers.

$(x - 3) (x - 2) = 0$

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

$(x - 3) (x - 2) = 0$

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

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

$x - 3 = 0$

$x - 2 = 0$

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

Set $x - 3$ equal to $0$ and solve for $x$ .

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

$x - 3 = 0$

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

Add $3$ to both sides of the equation.

$x = 3$

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

$x = 3$

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

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

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

$x - 2 = 0$

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

Add $2$ to both sides of the equation.

$x = 2$

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

$x = 2$

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

The final solution is all the values that make $(x - 3) (x - 2) = 0$ true.

$x = 3, 2$

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

$x = 3, 2$

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

Step 3

Replace all occurrences of $x$ with $3$ in each equation.

Replace all occurrences of $x$ in $y = x^{2} - 6 x + 9$ with $3$ .

$y = {(3)}^{2} - 6 \cdot 3 + 9$

$x = 3$

Simplify the right side.

Simplify ${(3)}^{2} - 6 \cdot 3 + 9$ .

Simplify each term.

Raise $3$ to the power of $2$ .

$y = 9 - 6 \cdot 3 + 9$

$x = 3$

Multiply $- 6$ by $3$ .

$y = 9 - 18 + 9$

$x = 3$

$y = 9 - 18 + 9$

$x = 3$

Simplify by adding and subtracting.

Subtract $18$ from $9$ .

$y = - 9 + 9$

$x = 3$

Add $- 9$ and $9$ .

$y = 0$

$x = 3$

$y = 0$

$x = 3$

$y = 0$

$x = 3$

$y = 0$

$x = 3$

$y = 0$

$x = 3$

Step 4

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

Replace all occurrences of $x$ in $y = x^{2} - 6 x + 9$ with $2$ .

$y = {(2)}^{2} - 6 \cdot 2 + 9$

$x = 2$

Simplify the right side.

Simplify ${(2)}^{2} - 6 \cdot 2 + 9$ .

Simplify each term.

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

$y = 4 - 6 \cdot 2 + 9$

$x = 2$

Multiply $- 6$ by $2$ .

$y = 4 - 12 + 9$

$x = 2$

$y = 4 - 12 + 9$

$x = 2$

Simplify by adding and subtracting.

Subtract $12$ from $4$ .

$y = - 8 + 9$

$x = 2$

Add $- 8$ and $9$ .

$y = 1$

$x = 2$

$y = 1$

$x = 2$

$y = 1$

$x = 2$

$y = 1$

$x = 2$

$y = 1$

$x = 2$

Step 5

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

$(3, 0)$

$(2, 1)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(3, 0), (2, 1)$

Equation Form:

$x = 3, y = 0$

$x = 2, y = 1$

Step 7

Solve the System of Equations x+y=3 y=x^2-6x+9

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