Solve the System of Equations x^2+y^2=400 y=x-28

$x^{2} + y^{2} = 400$ $y = x - 28$

Step 1

Replace all occurrences of $y$ with $x - 28$ in each equation.

Replace all occurrences of $y$ in $x^{2} + y^{2} = 400$ with $x - 28$ .

$x^{2} + {(x - 28)}^{2} = 400$

$y = x - 28$

Simplify the left side.

Simplify $x^{2} + {(x - 28)}^{2}$ .

Simplify each term.

Rewrite ${(x - 28)}^{2}$ as $(x - 28) (x - 28)$ .

$x^{2} + (x - 28) (x - 28) = 400$

$y = x - 28$

Expand $(x - 28) (x - 28)$ using the FOIL Method.

Apply the distributive property.

$x^{2} + x (x - 28) - 28 (x - 28) = 400$

$y = x - 28$

Apply the distributive property.

$x^{2} + x \cdot x + x \cdot - 28 - 28 (x - 28) = 400$

$y = x - 28$

Apply the distributive property.

$x^{2} + x \cdot x + x \cdot - 28 - 28 x - 28 \cdot - 28 = 400$

$y = x - 28$

$x^{2} + x \cdot x + x \cdot - 28 - 28 x - 28 \cdot - 28 = 400$

$y = x - 28$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + x^{2} + x \cdot - 28 - 28 x - 28 \cdot - 28 = 400$

$y = x - 28$

Move $- 28$ to the left of $x$ .

$x^{2} + x^{2} - 28 \cdot x - 28 x - 28 \cdot - 28 = 400$

$y = x - 28$

Multiply $- 28$ by $- 28$ .

$x^{2} + x^{2} - 28 x - 28 x + 784 = 400$

$y = x - 28$

$x^{2} + x^{2} - 28 x - 28 x + 784 = 400$

$y = x - 28$

Subtract $28 x$ from $- 28 x$ .

$x^{2} + x^{2} - 56 x + 784 = 400$

$y = x - 28$

$x^{2} + x^{2} - 56 x + 784 = 400$

$y = x - 28$

$x^{2} + x^{2} - 56 x + 784 = 400$

$y = x - 28$

Add $x^{2}$ and $x^{2}$ .

$2 x^{2} - 56 x + 784 = 400$

$y = x - 28$

$2 x^{2} - 56 x + 784 = 400$

$y = x - 28$

$2 x^{2} - 56 x + 784 = 400$

$y = x - 28$

$2 x^{2} - 56 x + 784 = 400$

$y = x - 28$

Step 2

Solve for $x$ in $2 x^{2} - 56 x + 784 = 400$ .

Subtract $400$ from both sides of the equation.

$2 x^{2} - 56 x + 784 - 400 = 0$

$y = x - 28$

Subtract $400$ from $784$ .

$2 x^{2} - 56 x + 384 = 0$

$y = x - 28$

Factor the left side of the equation.

Factor $2$ out of $2 x^{2} - 56 x + 384$ .

Factor $2$ out of $2 x^{2}$ .

$2 (x^{2}) - 56 x + 384 = 0$

$y = x - 28$

Factor $2$ out of $- 56 x$ .

$2 (x^{2}) + 2 (- 28 x) + 384 = 0$

$y = x - 28$

Factor $2$ out of $384$ .

$2 x^{2} + 2 (- 28 x) + 2 \cdot 192 = 0$

$y = x - 28$

Factor $2$ out of $2 x^{2} + 2 (- 28 x)$ .

$2 (x^{2} - 28 x) + 2 \cdot 192 = 0$

$y = x - 28$

Factor $2$ out of $2 (x^{2} - 28 x) + 2 \cdot 192$ .

$2 (x^{2} - 28 x + 192) = 0$

$y = x - 28$

$2 (x^{2} - 28 x + 192) = 0$

$y = x - 28$

Factor.

Factor $x^{2} - 28 x + 192$ 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 $192$ and whose sum is $- 28$ .

$- 16, - 12$

$y = x - 28$

Write the factored form using these integers.

$2 ((x - 16) (x - 12)) = 0$

$y = x - 28$

$2 ((x - 16) (x - 12)) = 0$

$y = x - 28$

Remove unnecessary parentheses.

$2 (x - 16) (x - 12) = 0$

$y = x - 28$

$2 (x - 16) (x - 12) = 0$

$y = x - 28$

$2 (x - 16) (x - 12) = 0$

$y = x - 28$

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

$x - 16 = 0$

$x - 12 = 0$

$y = x - 28$

Set $x - 16$ equal to $0$ and solve for $x$ .

Set $x - 16$ equal to $0$ .

$x - 16 = 0$

$y = x - 28$

Add $16$ to both sides of the equation.

$x = 16$

$y = x - 28$

$x = 16$

$y = x - 28$

Set $x - 12$ equal to $0$ and solve for $x$ .

Set $x - 12$ equal to $0$ .

$x - 12 = 0$

$y = x - 28$

Add $12$ to both sides of the equation.

$x = 12$

$y = x - 28$

$x = 12$

$y = x - 28$

The final solution is all the values that make $2 (x - 16) (x - 12) = 0$ true.

$x = 16, 12$

$y = x - 28$

$x = 16, 12$

$y = x - 28$

Step 3

Replace all occurrences of $x$ with $16$ in each equation.

Replace all occurrences of $x$ in $y = x - 28$ with $16$ .

$y = (16) - 28$

$x = 16$

Simplify the right side.

Subtract $28$ from $16$ .

$y = - 12$

$x = 16$

$y = - 12$

$x = 16$

$y = - 12$

$x = 16$

Step 4

Replace all occurrences of $x$ with $12$ in each equation.

Replace all occurrences of $x$ in $y = x - 28$ with $12$ .

$y = (12) - 28$

$x = 12$

Simplify the right side.

Subtract $28$ from $12$ .

$y = - 16$

$x = 12$

$y = - 16$

$x = 12$

$y = - 16$

$x = 12$

Step 5

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

$(16, - 12)$

$(12, - 16)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(16, - 12), (12, - 16)$

Equation Form:

$x = 16, y = - 12$

$x = 12, y = - 16$

Step 7

Solve the System of Equations x^2+y^2=400 y=x-28

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