Solve by Substitution x^2+y^2=26 , y-x=4

$x^{2} + y^{2} = 26$ , $y - x = 4$

Step 1

Add $x$ to both sides of the equation.

$y = 4 + x$

$x^{2} + y^{2} = 26$

Step 2

Replace all occurrences of $y$ with $4 + x$ in each equation.

Replace all occurrences of $y$ in $x^{2} + y^{2} = 26$ with $4 + x$ .

$x^{2} + {(4 + x)}^{2} = 26$

$y = 4 + x$

Simplify the left side.

Simplify $x^{2} + {(4 + x)}^{2}$ .

Simplify each term.

Rewrite ${(4 + x)}^{2}$ as $(4 + x) (4 + x)$ .

$x^{2} + (4 + x) (4 + x) = 26$

$y = 4 + x$

Expand $(4 + x) (4 + x)$ using the FOIL Method.

Apply the distributive property.

$x^{2} + 4 (4 + x) + x (4 + x) = 26$

$y = 4 + x$

Apply the distributive property.

$x^{2} + 4 \cdot 4 + 4 x + x (4 + x) = 26$

$y = 4 + x$

Apply the distributive property.

$x^{2} + 4 \cdot 4 + 4 x + x \cdot 4 + x \cdot x = 26$

$y = 4 + x$

$x^{2} + 4 \cdot 4 + 4 x + x \cdot 4 + x \cdot x = 26$

$y = 4 + x$

Simplify and combine like terms.

Simplify each term.

Multiply $4$ by $4$ .

$x^{2} + 16 + 4 x + x \cdot 4 + x \cdot x = 26$

$y = 4 + x$

Move $4$ to the left of $x$ .

$x^{2} + 16 + 4 x + 4 \cdot x + x \cdot x = 26$

$y = 4 + x$

Multiply $x$ by $x$ .

$x^{2} + 16 + 4 x + 4 x + x^{2} = 26$

$y = 4 + x$

$x^{2} + 16 + 4 x + 4 x + x^{2} = 26$

$y = 4 + x$

Add $4 x$ and $4 x$ .

$x^{2} + 16 + 8 x + x^{2} = 26$

$y = 4 + x$

$x^{2} + 16 + 8 x + x^{2} = 26$

$y = 4 + x$

$x^{2} + 16 + 8 x + x^{2} = 26$

$y = 4 + x$

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

$2 x^{2} + 16 + 8 x = 26$

$y = 4 + x$

$2 x^{2} + 16 + 8 x = 26$

$y = 4 + x$

$2 x^{2} + 16 + 8 x = 26$

$y = 4 + x$

$2 x^{2} + 16 + 8 x = 26$

$y = 4 + x$

Step 3

Solve for $x$ in $2 x^{2} + 16 + 8 x = 26$ .

Subtract $26$ from both sides of the equation.

$2 x^{2} + 16 + 8 x - 26 = 0$

$y = 4 + x$

Subtract $26$ from $16$ .

$2 x^{2} + 8 x - 10 = 0$

$y = 4 + x$

Factor the left side of the equation.

Factor $2$ out of $2 x^{2} + 8 x - 10$ .

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

$2 (x^{2}) + 8 x - 10 = 0$

$y = 4 + x$

Factor $2$ out of $8 x$ .

$2 (x^{2}) + 2 (4 x) - 10 = 0$

$y = 4 + x$

Factor $2$ out of $- 10$ .

$2 x^{2} + 2 (4 x) + 2 \cdot - 5 = 0$

$y = 4 + x$

Factor $2$ out of $2 x^{2} + 2 (4 x)$ .

$2 (x^{2} + 4 x) + 2 \cdot - 5 = 0$

$y = 4 + x$

Factor $2$ out of $2 (x^{2} + 4 x) + 2 \cdot - 5$ .

$2 (x^{2} + 4 x - 5) = 0$

$y = 4 + x$

$2 (x^{2} + 4 x - 5) = 0$

$y = 4 + x$

Factor.

Factor $x^{2} + 4 x - 5$ 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 $- 5$ and whose sum is $4$ .

$- 1, 5$

$y = 4 + x$

Write the factored form using these integers.

$2 ((x - 1) (x + 5)) = 0$

$y = 4 + x$

$2 ((x - 1) (x + 5)) = 0$

$y = 4 + x$

Remove unnecessary parentheses.

$2 (x - 1) (x + 5) = 0$

$y = 4 + x$

$2 (x - 1) (x + 5) = 0$

$y = 4 + x$

$2 (x - 1) (x + 5) = 0$

$y = 4 + x$

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

$x - 1 = 0$

$x + 5 = 0$

$y = 4 + x$

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

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

$x - 1 = 0$

$y = 4 + x$

Add $1$ to both sides of the equation.

$x = 1$

$y = 4 + x$

$x = 1$

$y = 4 + x$

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

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

$x + 5 = 0$

$y = 4 + x$

Subtract $5$ from both sides of the equation.

$x = - 5$

$y = 4 + x$

$x = - 5$

$y = 4 + x$

The final solution is all the values that make $2 (x - 1) (x + 5) = 0$ true.

$x = 1, - 5$

$y = 4 + x$

$x = 1, - 5$

$y = 4 + x$

Step 4

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

Replace all occurrences of $x$ in $y = 4 + x$ with $1$ .

$y = 4 + 1$

$x = 1$

Simplify $y = 4 + 1$ .

Simplify the left side.

Remove parentheses.

$y = 4 + 1$

$x = 1$

$y = 4 + 1$

$x = 1$

Simplify the right side.

Add $4$ and $1$ .

$y = 5$

$x = 1$

$y = 5$

$x = 1$

$y = 5$

$x = 1$

$y = 5$

$x = 1$

Step 5

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

Replace all occurrences of $x$ in $y = 4 + x$ with $- 5$ .

$y = 4 - 5$

$x = - 5$

Simplify $y = 4 - 5$ .

Simplify the left side.

Remove parentheses.

$y = 4 - 5$

$x = - 5$

$y = 4 - 5$

$x = - 5$

Simplify the right side.

Subtract $5$ from $4$ .

$y = - 1$

$x = - 5$

$y = - 1$

$x = - 5$

$y = - 1$

$x = - 5$

$y = - 1$

$x = - 5$

Step 6

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

$(1, 5)$

$(- 5, - 1)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(1, 5), (- 5, - 1)$

Equation Form:

$x = 1, y = 5$

$x = - 5, y = - 1$

Step 8

Solve by Substitution x^2+y^2=26 , y-x=4

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