Solve by Substitution x+y=1 , x^2+xy-y^2=-29

$x + y = 1$ , $x^{2} + x y - y^{2} = - 29$

Step 1

Subtract $y$ from both sides of the equation.

$x = 1 - y$

$x^{2} + x y - y^{2} = - 29$

Step 2

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

Replace all occurrences of $x$ in $x^{2} + x y - y^{2} = - 29$ with $1 - y$ .

${(1 - y)}^{2} + (1 - y) \cdot y - y^{2} = - 29$

$x = 1 - y$

Simplify the left side.

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

Simplify each term.

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

$(1 - y) (1 - y) + (1 - y) \cdot y - y^{2} = - 29$

$x = 1 - y$

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

Apply the distributive property.

$1 (1 - y) - y (1 - y) + (1 - y) \cdot y - y^{2} = - 29$

$x = 1 - y$

Apply the distributive property.

$1 \cdot 1 + 1 (- y) - y (1 - y) + (1 - y) \cdot y - y^{2} = - 29$

$x = 1 - y$

Apply the distributive property.

$1 \cdot 1 + 1 (- y) - y \cdot 1 - y (- y) + (1 - y) \cdot y - y^{2} = - 29$

$x = 1 - y$

$1 \cdot 1 + 1 (- y) - y \cdot 1 - y (- y) + (1 - y) \cdot y - y^{2} = - 29$

$x = 1 - y$

Simplify and combine like terms.

Simplify each term.

Multiply $1$ by $1$ .

$1 + 1 (- y) - y \cdot 1 - y (- y) + (1 - y) \cdot y - y^{2} = - 29$

$x = 1 - y$

Multiply $- y$ by $1$ .

$1 - y - y \cdot 1 - y (- y) + (1 - y) \cdot y - y^{2} = - 29$

$x = 1 - y$

Multiply $- 1$ by $1$ .

$1 - y - y - y (- y) + (1 - y) \cdot y - y^{2} = - 29$

$x = 1 - y$

Rewrite using the commutative property of multiplication.

$1 - y - y - 1 \cdot (- 1 y \cdot y) + (1 - y) \cdot y - y^{2} = - 29$

$x = 1 - y$

Multiply $y$ by $y$ by adding the exponents.

Move $y$ .

$1 - y - y - 1 \cdot (- 1 (y \cdot y)) + (1 - y) \cdot y - y^{2} = - 29$

$x = 1 - y$

Multiply $y$ by $y$ .

$1 - y - y - 1 \cdot (- 1 y^{2}) + (1 - y) \cdot y - y^{2} = - 29$

$x = 1 - y$

$1 - y - y - 1 \cdot (- 1 y^{2}) + (1 - y) \cdot y - y^{2} = - 29$

$x = 1 - y$

Multiply $- 1$ by $- 1$ .

$1 - y - y + 1 y^{2} + (1 - y) \cdot y - y^{2} = - 29$

$x = 1 - y$

Multiply $y^{2}$ by $1$ .

$1 - y - y + y^{2} + (1 - y) \cdot y - y^{2} = - 29$

$x = 1 - y$

$1 - y - y + y^{2} + (1 - y) \cdot y - y^{2} = - 29$

$x = 1 - y$

Subtract $y$ from $- y$ .

$1 - 2 y + y^{2} + (1 - y) \cdot y - y^{2} = - 29$

$x = 1 - y$

$1 - 2 y + y^{2} + (1 - y) \cdot y - y^{2} = - 29$

$x = 1 - y$

Apply the distributive property.

$1 - 2 y + y^{2} + 1 y - y \cdot y - y^{2} = - 29$

$x = 1 - y$

Multiply $y$ by $1$ .

$1 - 2 y + y^{2} + y - y \cdot y - y^{2} = - 29$

$x = 1 - y$

Multiply $y$ by $y$ by adding the exponents.

Move $y$ .

$1 - 2 y + y^{2} + y - (y \cdot y) - y^{2} = - 29$

$x = 1 - y$

Multiply $y$ by $y$ .

$1 - 2 y + y^{2} + y - y^{2} - y^{2} = - 29$

$x = 1 - y$

$1 - 2 y + y^{2} + y - y^{2} - y^{2} = - 29$

$x = 1 - y$

$1 - 2 y + y^{2} + y - y^{2} - y^{2} = - 29$

$x = 1 - y$

Simplify by adding terms.

Combine the opposite terms in $1 - 2 y + y^{2} + y - y^{2} - y^{2}$ .

Subtract $y^{2}$ from $y^{2}$ .

$1 - 2 y + y + 0 - y^{2} = - 29$

$x = 1 - y$

Add $1 - 2 y + y$ and $0$ .

$1 - 2 y + y - y^{2} = - 29$

$x = 1 - y$

$1 - 2 y + y - y^{2} = - 29$

$x = 1 - y$

Add $- 2 y$ and $y$ .

$1 - y - y^{2} = - 29$

$x = 1 - y$

$1 - y - y^{2} = - 29$

$x = 1 - y$

$1 - y - y^{2} = - 29$

$x = 1 - y$

$1 - y - y^{2} = - 29$

$x = 1 - y$

$1 - y - y^{2} = - 29$

$x = 1 - y$

Step 3

Solve for $y$ in $1 - y - y^{2} = - 29$ .

Add $29$ to both sides of the equation.

$1 - y - y^{2} + 29 = 0$

$x = 1 - y$

Add $1$ and $29$ .

$- y - y^{2} + 30 = 0$

$x = 1 - y$

Factor the left side of the equation.

Factor $- 1$ out of $- y - y^{2} + 30$ .

Reorder $- y$ and $- y^{2}$ .

$- y^{2} - y + 30 = 0$

$x = 1 - y$

Factor $- 1$ out of $- y^{2}$ .

$- (y^{2}) - y + 30 = 0$

$x = 1 - y$

Factor $- 1$ out of $- y$ .

$- (y^{2}) - (y) + 30 = 0$

$x = 1 - y$

Rewrite $30$ as $- 1 (- 30)$ .

$- (y^{2}) - (y) - 1 \cdot - 30 = 0$

$x = 1 - y$

Factor $- 1$ out of $- (y^{2}) - (y)$ .

$- (y^{2} + y) - 1 \cdot - 30 = 0$

$x = 1 - y$

Factor $- 1$ out of $- (y^{2} + y) - 1 (- 30)$ .

$- (y^{2} + y - 30) = 0$

$x = 1 - y$

$- (y^{2} + y - 30) = 0$

$x = 1 - y$

Factor.

Factor $y^{2} + y - 30$ 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 $- 30$ and whose sum is $1$ .

$- 5, 6$

$x = 1 - y$

Write the factored form using these integers.

$- ((y - 5) (y + 6)) = 0$

$x = 1 - y$

$- ((y - 5) (y + 6)) = 0$

$x = 1 - y$

Remove unnecessary parentheses.

$- (y - 5) (y + 6) = 0$

$x = 1 - y$

$- (y - 5) (y + 6) = 0$

$x = 1 - y$

$- (y - 5) (y + 6) = 0$

$x = 1 - y$

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

$y - 5 = 0$

$y + 6 = 0$

$x = 1 - y$

Set $y - 5$ equal to $0$ and solve for $y$ .

Set $y - 5$ equal to $0$ .

$y - 5 = 0$

$x = 1 - y$

Add $5$ to both sides of the equation.

$y = 5$

$x = 1 - y$

$y = 5$

$x = 1 - y$

Set $y + 6$ equal to $0$ and solve for $y$ .

Set $y + 6$ equal to $0$ .

$y + 6 = 0$

$x = 1 - y$

Subtract $6$ from both sides of the equation.

$y = - 6$

$x = 1 - y$

$y = - 6$

$x = 1 - y$

The final solution is all the values that make $- (y - 5) (y + 6) = 0$ true.

$y = 5, - 6$

$x = 1 - y$

$y = 5, - 6$

$x = 1 - y$

Step 4

Replace all occurrences of $y$ with $5$ in each equation.

Replace all occurrences of $y$ in $x = 1 - y$ with $5$ .

$x = 1 - (5)$

$y = 5$

Simplify the right side.

Simplify $1 - (5)$ .

Multiply $- 1$ by $5$ .

$x = 1 - 5$

$y = 5$

Subtract $5$ from $1$ .

$x = - 4$

$y = 5$

$x = - 4$

$y = 5$

$x = - 4$

$y = 5$

$x = - 4$

$y = 5$

Step 5

Replace all occurrences of $y$ with $- 6$ in each equation.

Replace all occurrences of $y$ in $x = 1 - y$ with $- 6$ .

$x = 1 - (- 6)$

$y = - 6$

Simplify the right side.

Simplify $1 - (- 6)$ .

Multiply $- 1$ by $- 6$ .

$x = 1 + 6$

$y = - 6$

Add $1$ and $6$ .

$x = 7$

$y = - 6$

$x = 7$

$y = - 6$

$x = 7$

$y = - 6$

$x = 7$

$y = - 6$

Step 6

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

$(- 4, 5)$

$(7, - 6)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(- 4, 5), (7, - 6)$

Equation Form:

$x = - 4, y = 5$

$x = 7, y = - 6$

Step 8

Solve by Substitution x+y=1 , x^2+xy-y^2=-29

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