Solve by Substitution x+y=3 , y=x^2-6x+7

$x + y = 3$ , $y = x^{2} - 6 x + 7$

Step 1

Replace all occurrences of $y$ with $x^{2} - 6 x + 7$ in each equation.

Replace all occurrences of $y$ in $x + y = 3$ with $x^{2} - 6 x + 7$ .

$x + x^{2} - 6 x + 7 = 3$

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

Simplify the left side.

Simplify $x + x^{2} - 6 x + 7$ .

Remove parentheses.

$x + x^{2} - 6 x + 7 = 3$

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

Subtract $6 x$ from $x$ .

$x^{2} - 5 x + 7 = 3$

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

$x^{2} - 5 x + 7 = 3$

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

$x^{2} - 5 x + 7 = 3$

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

$x^{2} - 5 x + 7 = 3$

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

Step 2

Solve for $x$ in $x^{2} - 5 x + 7 = 3$ .

Subtract $3$ from both sides of the equation.

$x^{2} - 5 x + 7 - 3 = 0$

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

Subtract $3$ from $7$ .

$x^{2} - 5 x + 4 = 0$

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

Factor $x^{2} - 5 x + 4$ 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 $4$ and whose sum is $- 5$ .

$- 4, - 1$

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

Write the factored form using these integers.

$(x - 4) (x - 1) = 0$

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

$(x - 4) (x - 1) = 0$

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

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

$x - 4 = 0$

$x - 1 = 0$

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

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

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

Add $4$ to both sides of the equation.

$x = 4$

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

$x = 4$

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

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

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

$x - 1 = 0$

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

Add $1$ to both sides of the equation.

$x = 1$

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

$x = 1$

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

The final solution is all the values that make $(x - 4) (x - 1) = 0$ true.

$x = 4, 1$

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

$x = 4, 1$

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

Step 3

Replace all occurrences of $x$ with $4$ in each equation.

Replace all occurrences of $x$ in $y = x^{2} - 6 x + 7$ with $4$ .

$y = {(4)}^{2} - 6 \cdot 4 + 7$

$x = 4$

Simplify the right side.

Simplify ${(4)}^{2} - 6 \cdot 4 + 7$ .

Simplify each term.

Raise $4$ to the power of $2$ .

$y = 16 - 6 \cdot 4 + 7$

$x = 4$

Multiply $- 6$ by $4$ .

$y = 16 - 24 + 7$

$x = 4$

$y = 16 - 24 + 7$

$x = 4$

Simplify by adding and subtracting.

Subtract $24$ from $16$ .

$y = - 8 + 7$

$x = 4$

Add $- 8$ and $7$ .

$y = - 1$

$x = 4$

$y = - 1$

$x = 4$

$y = - 1$

$x = 4$

$y = - 1$

$x = 4$

$y = - 1$

$x = 4$

Step 4

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

Replace all occurrences of $x$ in $y = x^{2} - 6 x + 7$ with $1$ .

$y = {(1)}^{2} - 6 \cdot 1 + 7$

$x = 1$

Simplify the right side.

Simplify ${(1)}^{2} - 6 \cdot 1 + 7$ .

Simplify each term.

One to any power is one.

$y = 1 - 6 \cdot 1 + 7$

$x = 1$

Multiply $- 6$ by $1$ .

$y = 1 - 6 + 7$

$x = 1$

$y = 1 - 6 + 7$

$x = 1$

Simplify by adding and subtracting.

Subtract $6$ from $1$ .

$y = - 5 + 7$

$x = 1$

Add $- 5$ and $7$ .

$y = 2$

$x = 1$

$y = 2$

$x = 1$

$y = 2$

$x = 1$

$y = 2$

$x = 1$

$y = 2$

$x = 1$

Step 5

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

$(4, - 1)$

$(1, 2)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(4, - 1), (1, 2)$

Equation Form:

$x = 4, y = - 1$

$x = 1, y = 2$

Step 7

Solve by Substitution x+y=3 , y=x^2-6x+7

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