Solve by Substitution x^2-y=3 , x-y=-3

$x^{2} - y = 3$ , $x - y = - 3$

Add $y$ to both sides of the equation.

$x = - 3 + y$

$x^{2} - y = 3$

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

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

${(- 3 + y)}^{2} - y = 3$

$x = - 3 + y$

Simplify ${(- 3 + y)}^{2} - y$ .

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Simplify each term.

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Rewrite ${(- 3 + y)}^{2}$ as $(- 3 + y) (- 3 + y)$ .

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

$x = - 3 + y$

Expand $(- 3 + y) (- 3 + y)$ using the FOIL Method.

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Apply the distributive property.

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

$x = - 3 + y$

Apply the distributive property.

$- 3 \cdot - 3 - 3 y + y (- 3 + y) - y = 3$

$x = - 3 + y$

Apply the distributive property.

$- 3 \cdot - 3 - 3 y + y \cdot - 3 + y \cdot y - y = 3$

$x = - 3 + y$

$- 3 \cdot - 3 - 3 y + y \cdot - 3 + y \cdot y - y = 3$

$x = - 3 + y$

Simplify and combine like terms.

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Simplify each term.

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Multiply $- 3$ by $- 3$ .

$9 - 3 y + y \cdot - 3 + y \cdot y - y = 3$

$x = - 3 + y$

Move $- 3$ to the left of $y$ .

$9 - 3 y - 3 \cdot y + y \cdot y - y = 3$

$x = - 3 + y$

Multiply $y$ by $y$ .

$9 - 3 y - 3 y + y^{2} - y = 3$

$x = - 3 + y$

$9 - 3 y - 3 y + y^{2} - y = 3$

$x = - 3 + y$

Subtract $3 y$ from $- 3 y$ .

$9 - 6 y + y^{2} - y = 3$

$x = - 3 + y$

$9 - 6 y + y^{2} - y = 3$

$x = - 3 + y$

$9 - 6 y + y^{2} - y = 3$

$x = - 3 + y$

Subtract $y$ from $- 6 y$ .

$9 + y^{2} - 7 y = 3$

$x = - 3 + y$

$9 + y^{2} - 7 y = 3$

$x = - 3 + y$

$9 + y^{2} - 7 y = 3$

$x = - 3 + y$

Solve for $y$ in the first equation.

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Subtract $3$ from both sides of the equation.

$9 + y^{2} - 7 y - 3 = 0$

$x = - 3 + y$

Subtract $3$ from $9$ .

$y^{2} - 7 y + 6 = 0$

$x = - 3 + y$

Factor $y^{2} - 7 y + 6$ using the AC method.

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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 $- 7$ .

$- 6, - 1$

$x = - 3 + y$

Write the factored form using these integers.

$(y - 6) (y - 1) = 0$

$x = - 3 + y$

$(y - 6) (y - 1) = 0$

$x = - 3 + y$

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

$y - 6 = 0$

$y - 1 = 0$

$x = - 3 + y$

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

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

$y - 6 = 0$

$x = - 3 + y$

Add $6$ to both sides of the equation.

$y = 6$

$x = - 3 + y$

$y = 6$

$x = - 3 + y$

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

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

$y - 1 = 0$

$x = - 3 + y$

Add $1$ to both sides of the equation.

$y = 1$

$x = - 3 + y$

$y = 1$

$x = - 3 + y$

The final solution is all the values that make $(y - 6) (y - 1) = 0$ true.

$y = 6, 1$

$x = - 3 + y$

$y = 6, 1$

$x = - 3 + y$

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

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Replace all occurrences of $y$ in $x = - 3 + y$ with $6$ .

$x = - 3 + 6$

$y = 6$

Simplify $x = - 3 + 6$ .

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Remove parentheses.

$x = - 3 + 6$

$y = 6$

Add $- 3$ and $6$ .

$x = 3$

$y = 6$

$x = 3$

$y = 6$

$x = 3$

$y = 6$

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

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

$x = - 3 + 1$

$y = 1$

Simplify $x = - 3 + 1$ .

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Remove parentheses.

$x = - 3 + 1$

$y = 1$

Add $- 3$ and $1$ .

$x = - 2$

$y = 1$

$x = - 2$

$y = 1$

$x = - 2$

$y = 1$

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

$(3, 6)$

$(- 2, 1)$

The result can be shown in multiple forms.

Point Form:

$(3, 6), (- 2, 1)$

Equation Form:

$x = 3, y = 6$

$x = - 2, y = 1$

Solve by Substitution x^2-y=3 , x-y=-3

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