Solve by Substitution x^2+y^2=89 , y-x=3

$x^{2} + y^{2} = 89$ , $y - x = 3$

Step 1

Add $x$ to both sides of the equation.

$y = 3 + x$

$x^{2} + y^{2} = 89$

Step 2

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

Replace all occurrences of $y$ in $x^{2} + y^{2} = 89$ with $3 + x$ .

$x^{2} + {(3 + x)}^{2} = 89$

$y = 3 + x$

Simplify the left side.

Simplify $x^{2} + {(3 + x)}^{2}$ .

Simplify each term.

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

$x^{2} + (3 + x) (3 + x) = 89$

$y = 3 + x$

Expand $(3 + x) (3 + x)$ using the FOIL Method.

Apply the distributive property.

$x^{2} + 3 (3 + x) + x (3 + x) = 89$

$y = 3 + x$

Apply the distributive property.

$x^{2} + 3 \cdot 3 + 3 x + x (3 + x) = 89$

$y = 3 + x$

Apply the distributive property.

$x^{2} + 3 \cdot 3 + 3 x + x \cdot 3 + x \cdot x = 89$

$y = 3 + x$

$x^{2} + 3 \cdot 3 + 3 x + x \cdot 3 + x \cdot x = 89$

$y = 3 + x$

Simplify and combine like terms.

Simplify each term.

Multiply $3$ by $3$ .

$x^{2} + 9 + 3 x + x \cdot 3 + x \cdot x = 89$

$y = 3 + x$

Move $3$ to the left of $x$ .

$x^{2} + 9 + 3 x + 3 \cdot x + x \cdot x = 89$

$y = 3 + x$

Multiply $x$ by $x$ .

$x^{2} + 9 + 3 x + 3 x + x^{2} = 89$

$y = 3 + x$

$x^{2} + 9 + 3 x + 3 x + x^{2} = 89$

$y = 3 + x$

Add $3 x$ and $3 x$ .

$x^{2} + 9 + 6 x + x^{2} = 89$

$y = 3 + x$

$x^{2} + 9 + 6 x + x^{2} = 89$

$y = 3 + x$

$x^{2} + 9 + 6 x + x^{2} = 89$

$y = 3 + x$

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

$2 x^{2} + 9 + 6 x = 89$

$y = 3 + x$

$2 x^{2} + 9 + 6 x = 89$

$y = 3 + x$

$2 x^{2} + 9 + 6 x = 89$

$y = 3 + x$

$2 x^{2} + 9 + 6 x = 89$

$y = 3 + x$

Step 3

Solve for $x$ in $2 x^{2} + 9 + 6 x = 89$ .

Subtract $89$ from both sides of the equation.

$2 x^{2} + 9 + 6 x - 89 = 0$

$y = 3 + x$

Subtract $89$ from $9$ .

$2 x^{2} + 6 x - 80 = 0$

$y = 3 + x$

Factor the left side of the equation.

Factor $2$ out of $2 x^{2} + 6 x - 80$ .

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

$2 (x^{2}) + 6 x - 80 = 0$

$y = 3 + x$

Factor $2$ out of $6 x$ .

$2 (x^{2}) + 2 (3 x) - 80 = 0$

$y = 3 + x$

Factor $2$ out of $- 80$ .

$2 x^{2} + 2 (3 x) + 2 \cdot - 40 = 0$

$y = 3 + x$

Factor $2$ out of $2 x^{2} + 2 (3 x)$ .

$2 (x^{2} + 3 x) + 2 \cdot - 40 = 0$

$y = 3 + x$

Factor $2$ out of $2 (x^{2} + 3 x) + 2 \cdot - 40$ .

$2 (x^{2} + 3 x - 40) = 0$

$y = 3 + x$

$2 (x^{2} + 3 x - 40) = 0$

$y = 3 + x$

Factor.

Factor $x^{2} + 3 x - 40$ 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 $- 40$ and whose sum is $3$ .

$- 5, 8$

$y = 3 + x$

Write the factored form using these integers.

$2 ((x - 5) (x + 8)) = 0$

$y = 3 + x$

$2 ((x - 5) (x + 8)) = 0$

$y = 3 + x$

Remove unnecessary parentheses.

$2 (x - 5) (x + 8) = 0$

$y = 3 + x$

$2 (x - 5) (x + 8) = 0$

$y = 3 + x$

$2 (x - 5) (x + 8) = 0$

$y = 3 + x$

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

$x - 5 = 0$

$x + 8 = 0$

$y = 3 + x$

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

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

$x - 5 = 0$

$y = 3 + x$

Add $5$ to both sides of the equation.

$x = 5$

$y = 3 + x$

$x = 5$

$y = 3 + x$

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

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

$x + 8 = 0$

$y = 3 + x$

Subtract $8$ from both sides of the equation.

$x = - 8$

$y = 3 + x$

$x = - 8$

$y = 3 + x$

The final solution is all the values that make $2 (x - 5) (x + 8) = 0$ true.

$x = 5, - 8$

$y = 3 + x$

$x = 5, - 8$

$y = 3 + x$

Step 4

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

Replace all occurrences of $x$ in $y = 3 + x$ with $5$ .

$y = 3 + 5$

$x = 5$

Simplify $y = 3 + 5$ .

Simplify the left side.

Remove parentheses.

$y = 3 + 5$

$x = 5$

$y = 3 + 5$

$x = 5$

Simplify the right side.

Add $3$ and $5$ .

$y = 8$

$x = 5$

$y = 8$

$x = 5$

$y = 8$

$x = 5$

$y = 8$

$x = 5$

Step 5

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

Replace all occurrences of $x$ in $y = 3 + x$ with $- 8$ .

$y = 3 - 8$

$x = - 8$

Simplify $y = 3 - 8$ .

Simplify the left side.

Remove parentheses.

$y = 3 - 8$

$x = - 8$

$y = 3 - 8$

$x = - 8$

Simplify the right side.

Subtract $8$ from $3$ .

$y = - 5$

$x = - 8$

$y = - 5$

$x = - 8$

$y = - 5$

$x = - 8$

$y = - 5$

$x = - 8$

Step 6

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

$(5, 8)$

$(- 8, - 5)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(5, 8), (- 8, - 5)$

Equation Form:

$x = 5, y = 8$

$x = - 8, y = - 5$

Step 8

Solve by Substitution x^2+y^2=89 , y-x=3

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