Solve by Substitution x+y=3 , y=x^2-5x+6

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

Step 1

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

Replace all occurrences of $y$ in $x + y = 3$ with $x^{2} - 5 x + 6$ .

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

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

Simplify the left side.

Simplify $x + x^{2} - 5 x + 6$ .

Remove parentheses.

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

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

Subtract $5 x$ from $x$ .

$x^{2} - 4 x + 6 = 3$

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

$x^{2} - 4 x + 6 = 3$

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

$x^{2} - 4 x + 6 = 3$

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

$x^{2} - 4 x + 6 = 3$

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

Step 2

Solve for $x$ in $x^{2} - 4 x + 6 = 3$ .

Subtract $3$ from both sides of the equation.

$x^{2} - 4 x + 6 - 3 = 0$

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

Subtract $3$ from $6$ .

$x^{2} - 4 x + 3 = 0$

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

Factor $x^{2} - 4 x + 3$ 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 $3$ and whose sum is $- 4$ .

$- 3, - 1$

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

Write the factored form using these integers.

$(x - 3) (x - 1) = 0$

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

$(x - 3) (x - 1) = 0$

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

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

$x - 3 = 0$

$x - 1 = 0$

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

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

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

$x - 3 = 0$

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

Add $3$ to both sides of the equation.

$x = 3$

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

$x = 3$

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

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

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

Add $1$ to both sides of the equation.

$x = 1$

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

$x = 1$

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

The final solution is all the values that make $(x - 3) (x - 1) = 0$ true.

$x = 3, 1$

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

$x = 3, 1$

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

Step 3

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

Replace all occurrences of $x$ in $y = x^{2} - 5 x + 6$ with $3$ .

$y = {(3)}^{2} - 5 \cdot 3 + 6$

$x = 3$

Simplify the right side.

Simplify ${(3)}^{2} - 5 \cdot 3 + 6$ .

Simplify each term.

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

$y = 9 - 5 \cdot 3 + 6$

$x = 3$

Multiply $- 5$ by $3$ .

$y = 9 - 15 + 6$

$x = 3$

$y = 9 - 15 + 6$

$x = 3$

Simplify by adding and subtracting.

Subtract $15$ from $9$ .

$y = - 6 + 6$

$x = 3$

Add $- 6$ and $6$ .

$y = 0$

$x = 3$

$y = 0$

$x = 3$

$y = 0$

$x = 3$

$y = 0$

$x = 3$

$y = 0$

$x = 3$

Step 4

Replace all occurrences of $x$ with $1$ in each equation.

Replace all occurrences of $x$ in $y = x^{2} - 5 x + 6$ with $1$ .

$y = {(1)}^{2} - 5 \cdot 1 + 6$

$x = 1$

Simplify the right side.

Simplify ${(1)}^{2} - 5 \cdot 1 + 6$ .

Simplify each term.

One to any power is one.

$y = 1 - 5 \cdot 1 + 6$

$x = 1$

Multiply $- 5$ by $1$ .

$y = 1 - 5 + 6$

$x = 1$

$y = 1 - 5 + 6$

$x = 1$

Simplify by adding and subtracting.

Subtract $5$ from $1$ .

$y = - 4 + 6$

$x = 1$

Add $- 4$ and $6$ .

$y = 2$

$x = 1$

$y = 2$

$x = 1$

$y = 2$

$x = 1$

$y = 2$

$x = 1$

$y = 2$

$x = 1$

Step 5

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

$(3, 0)$

$(1, 2)$

Step 6

The result can be shown in multiple forms.

Point Form:

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

Equation Form:

$x = 3, y = 0$

$x = 1, y = 2$

Step 7

Solve by Substitution x+y=3 , y=x^2-5x+6

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