Solve by Substitution x+y=2 , y=x^2-4x+4

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

Step 1

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

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

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

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

Simplify the left side.

Simplify $x + x^{2} - 4 x + 4$ .

Remove parentheses.

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

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

Subtract $4 x$ from $x$ .

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

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

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

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

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

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

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

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

Step 2

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

Subtract $2$ from both sides of the equation.

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

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

Subtract $2$ from $4$ .

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

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

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

$- 2, - 1$

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

Write the factored form using these integers.

$(x - 2) (x - 1) = 0$

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

$(x - 2) (x - 1) = 0$

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

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

$x - 2 = 0$

$x - 1 = 0$

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

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

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

$x - 2 = 0$

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

Add $2$ to both sides of the equation.

$x = 2$

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

$x = 2$

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

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

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

$x - 1 = 0$

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

Add $1$ to both sides of the equation.

$x = 1$

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

$x = 1$

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

The final solution is all the values that make $(x - 2) (x - 1) = 0$ true.

$x = 2, 1$

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

$x = 2, 1$

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

Step 3

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

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

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

$x = 2$

Simplify the right side.

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

Simplify each term.

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

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

$x = 2$

Multiply $- 4$ by $2$ .

$y = 4 - 8 + 4$

$x = 2$

$y = 4 - 8 + 4$

$x = 2$

Simplify by adding and subtracting.

Subtract $8$ from $4$ .

$y = - 4 + 4$

$x = 2$

Add $- 4$ and $4$ .

$y = 0$

$x = 2$

$y = 0$

$x = 2$

$y = 0$

$x = 2$

$y = 0$

$x = 2$

$y = 0$

$x = 2$

Step 4

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

Replace all occurrences of $x$ in $y = x^{2} - 4 x + 4$ with $1$ .

$y = {(1)}^{2} - 4 \cdot 1 + 4$

$x = 1$

Simplify the right side.

Simplify ${(1)}^{2} - 4 \cdot 1 + 4$ .

Simplify each term.

One to any power is one.

$y = 1 - 4 \cdot 1 + 4$

$x = 1$

Multiply $- 4$ by $1$ .

$y = 1 - 4 + 4$

$x = 1$

$y = 1 - 4 + 4$

$x = 1$

Simplify by adding and subtracting.

Subtract $4$ from $1$ .

$y = - 3 + 4$

$x = 1$

Add $- 3$ and $4$ .

$y = 1$

$x = 1$

$y = 1$

$x = 1$

$y = 1$

$x = 1$

$y = 1$

$x = 1$

$y = 1$

$x = 1$

Step 5

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

$(2, 0)$

$(1, 1)$

Step 6

The result can be shown in multiple forms.

Point Form:

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

Equation Form:

$x = 2, y = 0$

$x = 1, y = 1$

Step 7

Solve by Substitution x+y=2 , y=x^2-4x+4

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