Solve by Substitution x+y=3 , y=x^2-7x+11

$x + y = 3$ , $y = x^{2} - 7 x + 11$

Step 1

Replace all occurrences of $y$ with $x^{2} - 7 x + 11$ in each equation.

Replace all occurrences of $y$ in $x + y = 3$ with $x^{2} - 7 x + 11$ .

$x + x^{2} - 7 x + 11 = 3$

$y = x^{2} - 7 x + 11$

Simplify the left side.

Simplify $x + x^{2} - 7 x + 11$ .

Remove parentheses.

$x + x^{2} - 7 x + 11 = 3$

$y = x^{2} - 7 x + 11$

Subtract $7 x$ from $x$ .

$x^{2} - 6 x + 11 = 3$

$y = x^{2} - 7 x + 11$

$x^{2} - 6 x + 11 = 3$

$y = x^{2} - 7 x + 11$

$x^{2} - 6 x + 11 = 3$

$y = x^{2} - 7 x + 11$

$x^{2} - 6 x + 11 = 3$

$y = x^{2} - 7 x + 11$

Step 2

Solve for $x$ in $x^{2} - 6 x + 11 = 3$ .

Subtract $3$ from both sides of the equation.

$x^{2} - 6 x + 11 - 3 = 0$

$y = x^{2} - 7 x + 11$

Subtract $3$ from $11$ .

$x^{2} - 6 x + 8 = 0$

$y = x^{2} - 7 x + 11$

Factor $x^{2} - 6 x + 8$ 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 $8$ and whose sum is $- 6$ .

$- 4, - 2$

$y = x^{2} - 7 x + 11$

Write the factored form using these integers.

$(x - 4) (x - 2) = 0$

$y = x^{2} - 7 x + 11$

$(x - 4) (x - 2) = 0$

$y = x^{2} - 7 x + 11$

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

$x - 4 = 0$

$x - 2 = 0$

$y = x^{2} - 7 x + 11$

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

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

$x - 4 = 0$

$y = x^{2} - 7 x + 11$

Add $4$ to both sides of the equation.

$x = 4$

$y = x^{2} - 7 x + 11$

$x = 4$

$y = x^{2} - 7 x + 11$

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

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

$x - 2 = 0$

$y = x^{2} - 7 x + 11$

Add $2$ to both sides of the equation.

$x = 2$

$y = x^{2} - 7 x + 11$

$x = 2$

$y = x^{2} - 7 x + 11$

The final solution is all the values that make $(x - 4) (x - 2) = 0$ true.

$x = 4, 2$

$y = x^{2} - 7 x + 11$

$x = 4, 2$

$y = x^{2} - 7 x + 11$

Step 3

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

Replace all occurrences of $x$ in $y = x^{2} - 7 x + 11$ with $4$ .

$y = {(4)}^{2} - 7 \cdot 4 + 11$

$x = 4$

Simplify the right side.

Simplify ${(4)}^{2} - 7 \cdot 4 + 11$ .

Simplify each term.

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

$y = 16 - 7 \cdot 4 + 11$

$x = 4$

Multiply $- 7$ by $4$ .

$y = 16 - 28 + 11$

$x = 4$

$y = 16 - 28 + 11$

$x = 4$

Simplify by adding and subtracting.

Subtract $28$ from $16$ .

$y = - 12 + 11$

$x = 4$

Add $- 12$ and $11$ .

$y = - 1$

$x = 4$

$y = - 1$

$x = 4$

$y = - 1$

$x = 4$

$y = - 1$

$x = 4$

$y = - 1$

$x = 4$

Step 4

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

Replace all occurrences of $x$ in $y = x^{2} - 7 x + 11$ with $2$ .

$y = {(2)}^{2} - 7 \cdot 2 + 11$

$x = 2$

Simplify the right side.

Simplify ${(2)}^{2} - 7 \cdot 2 + 11$ .

Simplify each term.

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

$y = 4 - 7 \cdot 2 + 11$

$x = 2$

Multiply $- 7$ by $2$ .

$y = 4 - 14 + 11$

$x = 2$

$y = 4 - 14 + 11$

$x = 2$

Simplify by adding and subtracting.

Subtract $14$ from $4$ .

$y = - 10 + 11$

$x = 2$

Add $- 10$ and $11$ .

$y = 1$

$x = 2$

$y = 1$

$x = 2$

$y = 1$

$x = 2$

$y = 1$

$x = 2$

$y = 1$

$x = 2$

Step 5

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

$(4, - 1)$

$(2, 1)$

Step 6

The result can be shown in multiple forms.

Point Form:

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

Equation Form:

$x = 4, y = - 1$

$x = 2, y = 1$

Step 7

Solve by Substitution x+y=3 , y=x^2-7x+11

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