Solve by Substitution y^2+x^2=53 , y-x=5

$y^{2} + x^{2} = 53$ , $y - x = 5$

Step 1

Add $x$ to both sides of the equation.

$y = 5 + x$

$y^{2} + x^{2} = 53$

Step 2

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

Replace all occurrences of $y$ in $y^{2} + x^{2} = 53$ with $5 + x$ .

${(5 + x)}^{2} + x^{2} = 53$

$y = 5 + x$

Simplify the left side.

Simplify ${(5 + x)}^{2} + x^{2}$ .

Simplify each term.

Rewrite ${(5 + x)}^{2}$ as $(5 + x) (5 + x)$ .

$(5 + x) (5 + x) + x^{2} = 53$

$y = 5 + x$

Expand $(5 + x) (5 + x)$ using the FOIL Method.

Apply the distributive property.

$5 (5 + x) + x (5 + x) + x^{2} = 53$

$y = 5 + x$

Apply the distributive property.

$5 \cdot 5 + 5 x + x (5 + x) + x^{2} = 53$

$y = 5 + x$

Apply the distributive property.

$5 \cdot 5 + 5 x + x \cdot 5 + x \cdot x + x^{2} = 53$

$y = 5 + x$

$5 \cdot 5 + 5 x + x \cdot 5 + x \cdot x + x^{2} = 53$

$y = 5 + x$

Simplify and combine like terms.

Simplify each term.

Multiply $5$ by $5$ .

$25 + 5 x + x \cdot 5 + x \cdot x + x^{2} = 53$

$y = 5 + x$

Move $5$ to the left of $x$ .

$25 + 5 x + 5 \cdot x + x \cdot x + x^{2} = 53$

$y = 5 + x$

Multiply $x$ by $x$ .

$25 + 5 x + 5 x + x^{2} + x^{2} = 53$

$y = 5 + x$

$25 + 5 x + 5 x + x^{2} + x^{2} = 53$

$y = 5 + x$

Add $5 x$ and $5 x$ .

$25 + 10 x + x^{2} + x^{2} = 53$

$y = 5 + x$

$25 + 10 x + x^{2} + x^{2} = 53$

$y = 5 + x$

$25 + 10 x + x^{2} + x^{2} = 53$

$y = 5 + x$

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

$25 + 10 x + 2 x^{2} = 53$

$y = 5 + x$

$25 + 10 x + 2 x^{2} = 53$

$y = 5 + x$

$25 + 10 x + 2 x^{2} = 53$

$y = 5 + x$

$25 + 10 x + 2 x^{2} = 53$

$y = 5 + x$

Step 3

Solve for $x$ in $25 + 10 x + 2 x^{2} = 53$ .

Subtract $53$ from both sides of the equation.

$25 + 10 x + 2 x^{2} - 53 = 0$

$y = 5 + x$

Subtract $53$ from $25$ .

$10 x + 2 x^{2} - 28 = 0$

$y = 5 + x$

Factor the left side of the equation.

Factor $2$ out of $10 x + 2 x^{2} - 28$ .

Factor $2$ out of $10 x$ .

$2 (5 x) + 2 x^{2} - 28 = 0$

$y = 5 + x$

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

$2 (5 x) + 2 (x^{2}) - 28 = 0$

$y = 5 + x$

Factor $2$ out of $- 28$ .

$2 (5 x) + 2 x^{2} + 2 \cdot - 14 = 0$

$y = 5 + x$

Factor $2$ out of $2 (5 x) + 2 x^{2}$ .

$2 (5 x + x^{2}) + 2 \cdot - 14 = 0$

$y = 5 + x$

Factor $2$ out of $2 (5 x + x^{2}) + 2 \cdot - 14$ .

$2 (5 x + x^{2} - 14) = 0$

$y = 5 + x$

$2 (5 x + x^{2} - 14) = 0$

$y = 5 + x$

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

$2 (5 u + u^{2} - 14) = 0$

$y = 5 + x$

Factor $5 u + u^{2} - 14$ 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 $- 14$ and whose sum is $5$ .

$- 2, 7$

$y = 5 + x$

Write the factored form using these integers.

$2 ((u - 2) (u + 7)) = 0$

$y = 5 + x$

$2 ((u - 2) (u + 7)) = 0$

$y = 5 + x$

Factor.

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

$2 ((x - 2) (x + 7)) = 0$

$y = 5 + x$

Remove unnecessary parentheses.

$2 (x - 2) (x + 7) = 0$

$y = 5 + x$

$2 (x - 2) (x + 7) = 0$

$y = 5 + x$

$2 (x - 2) (x + 7) = 0$

$y = 5 + x$

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

$x - 2 = 0$

$x + 7 = 0$

$y = 5 + x$

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

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

$x - 2 = 0$

$y = 5 + x$

Add $2$ to both sides of the equation.

$x = 2$

$y = 5 + x$

$x = 2$

$y = 5 + x$

Set $x + 7$ equal to $0$ and solve for $x$ .

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

$y = 5 + x$

Subtract $7$ from both sides of the equation.

$x = - 7$

$y = 5 + x$

$x = - 7$

$y = 5 + x$

The final solution is all the values that make $2 (x - 2) (x + 7) = 0$ true.

$x = 2, - 7$

$y = 5 + x$

$x = 2, - 7$

$y = 5 + x$

Step 4

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

Replace all occurrences of $x$ in $y = 5 + x$ with $2$ .

$y = 5 + 2$

$x = 2$

Simplify $y = 5 + 2$ .

Simplify the left side.

Remove parentheses.

$y = 5 + 2$

$x = 2$

$y = 5 + 2$

$x = 2$

Simplify the right side.

Add $5$ and $2$ .

$y = 7$

$x = 2$

$y = 7$

$x = 2$

$y = 7$

$x = 2$

$y = 7$

$x = 2$

Step 5

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

Replace all occurrences of $x$ in $y = 5 + x$ with $- 7$ .

$y = 5 - 7$

$x = - 7$

Simplify $y = 5 - 7$ .

Simplify the left side.

Remove parentheses.

$y = 5 - 7$

$x = - 7$

$y = 5 - 7$

$x = - 7$

Simplify the right side.

Subtract $7$ from $5$ .

$y = - 2$

$x = - 7$

$y = - 2$

$x = - 7$

$y = - 2$

$x = - 7$

$y = - 2$

$x = - 7$

Step 6

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

$(2, 7)$

$(- 7, - 2)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(2, 7), (- 7, - 2)$

Equation Form:

$x = 2, y = 7$

$x = - 7, y = - 2$

Step 8

Solve by Substitution y^2+x^2=53 , y-x=5

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