Solve by Substitution x^2+y^2=13 , x-y=1

$x^{2} + y^{2} = 13$ , $x - y = 1$

Step 1

Add $y$ to both sides of the equation.

$x = 1 + y$

$x^{2} + y^{2} = 13$

Step 2

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

Replace all occurrences of $x$ in $x^{2} + y^{2} = 13$ with $1 + y$ .

${(1 + y)}^{2} + y^{2} = 13$

$x = 1 + y$

Simplify the left side.

Simplify ${(1 + y)}^{2} + y^{2}$ .

Simplify each term.

Rewrite ${(1 + y)}^{2}$ as $(1 + y) (1 + y)$ .

$(1 + y) (1 + y) + y^{2} = 13$

$x = 1 + y$

Expand $(1 + y) (1 + y)$ using the FOIL Method.

Apply the distributive property.

$1 (1 + y) + y (1 + y) + y^{2} = 13$

$x = 1 + y$

Apply the distributive property.

$1 \cdot 1 + 1 y + y (1 + y) + y^{2} = 13$

$x = 1 + y$

Apply the distributive property.

$1 \cdot 1 + 1 y + y \cdot 1 + y \cdot y + y^{2} = 13$

$x = 1 + y$

$1 \cdot 1 + 1 y + y \cdot 1 + y \cdot y + y^{2} = 13$

$x = 1 + y$

Simplify and combine like terms.

Simplify each term.

Multiply $1$ by $1$ .

$1 + 1 y + y \cdot 1 + y \cdot y + y^{2} = 13$

$x = 1 + y$

Multiply $y$ by $1$ .

$1 + y + y \cdot 1 + y \cdot y + y^{2} = 13$

$x = 1 + y$

Multiply $y$ by $1$ .

$1 + y + y + y \cdot y + y^{2} = 13$

$x = 1 + y$

Multiply $y$ by $y$ .

$1 + y + y + y^{2} + y^{2} = 13$

$x = 1 + y$

$1 + y + y + y^{2} + y^{2} = 13$

$x = 1 + y$

Add $y$ and $y$ .

$1 + 2 y + y^{2} + y^{2} = 13$

$x = 1 + y$

$1 + 2 y + y^{2} + y^{2} = 13$

$x = 1 + y$

$1 + 2 y + y^{2} + y^{2} = 13$

$x = 1 + y$

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

$1 + 2 y + 2 y^{2} = 13$

$x = 1 + y$

$1 + 2 y + 2 y^{2} = 13$

$x = 1 + y$

$1 + 2 y + 2 y^{2} = 13$

$x = 1 + y$

$1 + 2 y + 2 y^{2} = 13$

$x = 1 + y$

Step 3

Solve for $y$ in $1 + 2 y + 2 y^{2} = 13$ .

Subtract $13$ from both sides of the equation.

$1 + 2 y + 2 y^{2} - 13 = 0$

$x = 1 + y$

Subtract $13$ from $1$ .

$2 y + 2 y^{2} - 12 = 0$

$x = 1 + y$

Factor the left side of the equation.

Factor $2$ out of $2 y + 2 y^{2} - 12$ .

Factor $2$ out of $2 y$ .

$2 (y) + 2 y^{2} - 12 = 0$

$x = 1 + y$

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

$2 (y) + 2 (y^{2}) - 12 = 0$

$x = 1 + y$

Factor $2$ out of $- 12$ .

$2 (y) + 2 y^{2} + 2 \cdot - 6 = 0$

$x = 1 + y$

Factor $2$ out of $2 (y) + 2 y^{2}$ .

$2 (y + y^{2}) + 2 \cdot - 6 = 0$

$x = 1 + y$

Factor $2$ out of $2 (y + y^{2}) + 2 \cdot - 6$ .

$2 (y + y^{2} - 6) = 0$

$x = 1 + y$

$2 (y + y^{2} - 6) = 0$

$x = 1 + y$

Let $u = y$ . Substitute $u$ for all occurrences of $y$ .

$2 (u + u^{2} - 6) = 0$

$x = 1 + y$

Factor $u + u^{2} - 6$ 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 $- 6$ and whose sum is $1$ .

$- 2, 3$

$x = 1 + y$

Write the factored form using these integers.

$2 ((u - 2) (u + 3)) = 0$

$x = 1 + y$

$2 ((u - 2) (u + 3)) = 0$

$x = 1 + y$

Factor.

Replace all occurrences of $u$ with $y$ .

$2 ((y - 2) (y + 3)) = 0$

$x = 1 + y$

Remove unnecessary parentheses.

$2 (y - 2) (y + 3) = 0$

$x = 1 + y$

$2 (y - 2) (y + 3) = 0$

$x = 1 + y$

$2 (y - 2) (y + 3) = 0$

$x = 1 + y$

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

$y - 2 = 0$

$y + 3 = 0$

$x = 1 + y$

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

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

$y - 2 = 0$

$x = 1 + y$

Add $2$ to both sides of the equation.

$y = 2$

$x = 1 + y$

$y = 2$

$x = 1 + y$

Set $y + 3$ equal to $0$ and solve for $y$ .

Set $y + 3$ equal to $0$ .

$y + 3 = 0$

$x = 1 + y$

Subtract $3$ from both sides of the equation.

$y = - 3$

$x = 1 + y$

$y = - 3$

$x = 1 + y$

The final solution is all the values that make $2 (y - 2) (y + 3) = 0$ true.

$y = 2, - 3$

$x = 1 + y$

$y = 2, - 3$

$x = 1 + y$

Step 4

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

Replace all occurrences of $y$ in $x = 1 + y$ with $2$ .

$x = 1 + 2$

$y = 2$

Simplify $x = 1 + 2$ .

Simplify the left side.

Remove parentheses.

$x = 1 + 2$

$y = 2$

$x = 1 + 2$

$y = 2$

Simplify the right side.

Add $1$ and $2$ .

$x = 3$

$y = 2$

$x = 3$

$y = 2$

$x = 3$

$y = 2$

$x = 3$

$y = 2$

Step 5

Replace all occurrences of $y$ with $- 3$ in each equation.

Replace all occurrences of $y$ in $x = 1 + y$ with $- 3$ .

$x = 1 - 3$

$y = - 3$

Simplify $x = 1 - 3$ .

Simplify the left side.

Remove parentheses.

$x = 1 - 3$

$y = - 3$

$x = 1 - 3$

$y = - 3$

Simplify the right side.

Subtract $3$ from $1$ .

$x = - 2$

$y = - 3$

$x = - 2$

$y = - 3$

$x = - 2$

$y = - 3$

$x = - 2$

$y = - 3$

Step 6

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

$(3, 2)$

$(- 2, - 3)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(3, 2), (- 2, - 3)$

Equation Form:

$x = 3, y = 2$

$x = - 2, y = - 3$

Step 8

Solve by Substitution x^2+y^2=13 , x-y=1

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