Solve by Substitution y^2+x^2=65 , y+x=7

$y^{2} + x^{2} = 65$ , $y + x = 7$

Subtract $x$ from both sides of the equation.

$y = 7 - x$

$y^{2} + x^{2} = 65$

Replace all occurrences of $y$ with $7 - x$ in each equation.

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Replace all occurrences of $y$ in $y^{2} + x^{2} = 65$ with $7 - x$ .

${(7 - x)}^{2} + x^{2} = 65$

$y = 7 - x$

Simplify ${(7 - x)}^{2} + x^{2}$ .

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Simplify each term.

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Rewrite ${(7 - x)}^{2}$ as $(7 - x) (7 - x)$ .

$(7 - x) (7 - x) + x^{2} = 65$

$y = 7 - x$

Expand $(7 - x) (7 - x)$ using the FOIL Method.

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Apply the distributive property.

$7 (7 - x) - x (7 - x) + x^{2} = 65$

$y = 7 - x$

Apply the distributive property.

$7 \cdot 7 + 7 (- x) - x (7 - x) + x^{2} = 65$

$y = 7 - x$

Apply the distributive property.

$7 \cdot 7 + 7 (- x) - x \cdot 7 - x (- x) + x^{2} = 65$

$y = 7 - x$

$7 \cdot 7 + 7 (- x) - x \cdot 7 - x (- x) + x^{2} = 65$

$y = 7 - x$

Simplify and combine like terms.

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Simplify each term.

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Multiply $7$ by $7$ .

$49 + 7 (- x) - x \cdot 7 - x (- x) + x^{2} = 65$

$y = 7 - x$

Multiply $- 1$ by $7$ .

$49 - 7 x - x \cdot 7 - x (- x) + x^{2} = 65$

$y = 7 - x$

Multiply $7$ by $- 1$ .

$49 - 7 x - 7 x - x (- x) + x^{2} = 65$

$y = 7 - x$

Rewrite using the commutative property of multiplication.

$49 - 7 x - 7 x - 1 \cdot (- 1 x \cdot x) + x^{2} = 65$

$y = 7 - x$

Multiply $x$ by $x$ by adding the exponents.

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Move $x$ .

$49 - 7 x - 7 x - 1 \cdot (- 1 (x \cdot x)) + x^{2} = 65$

$y = 7 - x$

Multiply $x$ by $x$ .

$49 - 7 x - 7 x - 1 \cdot (- 1 x^{2}) + x^{2} = 65$

$y = 7 - x$

$49 - 7 x - 7 x - 1 \cdot (- 1 x^{2}) + x^{2} = 65$

$y = 7 - x$

Multiply $- 1$ by $- 1$ .

$49 - 7 x - 7 x + 1 x^{2} + x^{2} = 65$

$y = 7 - x$

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

$49 - 7 x - 7 x + x^{2} + x^{2} = 65$

$y = 7 - x$

$49 - 7 x - 7 x + x^{2} + x^{2} = 65$

$y = 7 - x$

Subtract $7 x$ from $- 7 x$ .

$49 - 14 x + x^{2} + x^{2} = 65$

$y = 7 - x$

$49 - 14 x + x^{2} + x^{2} = 65$

$y = 7 - x$

$49 - 14 x + x^{2} + x^{2} = 65$

$y = 7 - x$

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

$49 - 14 x + 2 x^{2} = 65$

$y = 7 - x$

$49 - 14 x + 2 x^{2} = 65$

$y = 7 - x$

$49 - 14 x + 2 x^{2} = 65$

$y = 7 - x$

Solve for $x$ in the first equation.

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Subtract $65$ from both sides of the equation.

$49 - 14 x + 2 x^{2} - 65 = 0$

$y = 7 - x$

Subtract $65$ from $49$ .

$- 14 x + 2 x^{2} - 16 = 0$

$y = 7 - x$

Factor the left side of the equation.

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Factor $2$ out of $- 14 x + 2 x^{2} - 16$ .

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Factor $2$ out of $- 14 x$ .

$2 (- 7 x) + 2 x^{2} - 16 = 0$

$y = 7 - x$

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

$2 (- 7 x) + 2 (x^{2}) - 16 = 0$

$y = 7 - x$

Factor $2$ out of $- 16$ .

$2 (- 7 x) + 2 x^{2} + 2 \cdot - 8 = 0$

$y = 7 - x$

Factor $2$ out of $2 (- 7 x) + 2 x^{2}$ .

$2 (- 7 x + x^{2}) + 2 \cdot - 8 = 0$

$y = 7 - x$

Factor $2$ out of $2 (- 7 x + x^{2}) + 2 \cdot - 8$ .

$2 (- 7 x + x^{2} - 8) = 0$

$y = 7 - x$

$2 (- 7 x + x^{2} - 8) = 0$

$y = 7 - x$

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

$2 (- 7 u + u^{2} - 8) = 0$

$y = 7 - x$

Factor $- 7 u + u^{2} - 8$ 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 $- 8$ and whose sum is $- 7$ .

$- 8, 1$

$y = 7 - x$

Write the factored form using these integers.

$2 ((u - 8) (u + 1)) = 0$

$y = 7 - x$

$2 ((u - 8) (u + 1)) = 0$

$y = 7 - x$

Factor.

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

$2 ((x - 8) (x + 1)) = 0$

$y = 7 - x$

Remove unnecessary parentheses.

$2 (x - 8) (x + 1) = 0$

$y = 7 - x$

$2 (x - 8) (x + 1) = 0$

$y = 7 - x$

$2 (x - 8) (x + 1) = 0$

$y = 7 - x$

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

$x - 8 = 0$

$x + 1 = 0$

$y = 7 - x$

Set $x - 8$ equal to $0$ and solve for $x$ .

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Set $x - 8$ equal to $0$ .

$x - 8 = 0$

$y = 7 - x$

Add $8$ to both sides of the equation.

$x = 8$

$y = 7 - x$

$x = 8$

$y = 7 - x$

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

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

$x + 1 = 0$

$y = 7 - x$

Subtract $1$ from both sides of the equation.

$x = - 1$

$y = 7 - x$

$x = - 1$

$y = 7 - x$

The final solution is all the values that make $2 (x - 8) (x + 1) = 0$ true.

$x = 8, - 1$

$y = 7 - x$

$x = 8, - 1$

$y = 7 - x$

Replace all occurrences of $x$ with $8$ in each equation.

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Replace all occurrences of $x$ in $y = 7 - x$ with $8$ .

$y = 7 - (8)$

$x = 8$

Simplify $7 - (8)$ .

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Multiply $- 1$ by $8$ .

$y = 7 - 8$

$x = 8$

Subtract $8$ from $7$ .

$y = - 1$

$x = 8$

$y = - 1$

$x = 8$

$y = - 1$

$x = 8$

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

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Replace all occurrences of $x$ in $y = 7 - x$ with $- 1$ .

$y = 7 - (- 1)$

$x = - 1$

Simplify $7 - (- 1)$ .

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Multiply $- 1$ by $- 1$ .

$y = 7 + 1$

$x = - 1$

Add $7$ and $1$ .

$y = 8$

$x = - 1$

$y = 8$

$x = - 1$

$y = 8$

$x = - 1$

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

$(8, - 1)$

$(- 1, 8)$

The result can be shown in multiple forms.

Point Form:

$(8, - 1), (- 1, 8)$

Equation Form:

$x = 8, y = - 1$

$x = - 1, y = 8$

Solve by Substitution y^2+x^2=65 , y+x=7

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