Solve by Substitution x^2+y^2=25 , y=x+1

$x^{2} + y^{2} = 25$ , $y = x + 1$

Step 1

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

Replace all occurrences of $y$ in $x^{2} + y^{2} = 25$ with $x + 1$ .

$x^{2} + {(x + 1)}^{2} = 25$

$y = x + 1$

Simplify the left side.

Simplify $x^{2} + {(x + 1)}^{2}$ .

Simplify each term.

Rewrite ${(x + 1)}^{2}$ as $(x + 1) (x + 1)$ .

$x^{2} + (x + 1) (x + 1) = 25$

$y = x + 1$

Expand $(x + 1) (x + 1)$ using the FOIL Method.

Apply the distributive property.

$x^{2} + x (x + 1) + 1 (x + 1) = 25$

$y = x + 1$

Apply the distributive property.

$x^{2} + x \cdot x + x \cdot 1 + 1 (x + 1) = 25$

$y = x + 1$

Apply the distributive property.

$x^{2} + x \cdot x + x \cdot 1 + 1 x + 1 \cdot 1 = 25$

$y = x + 1$

$x^{2} + x \cdot x + x \cdot 1 + 1 x + 1 \cdot 1 = 25$

$y = x + 1$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + x^{2} + x \cdot 1 + 1 x + 1 \cdot 1 = 25$

$y = x + 1$

Multiply $x$ by $1$ .

$x^{2} + x^{2} + x + 1 x + 1 \cdot 1 = 25$

$y = x + 1$

Multiply $x$ by $1$ .

$x^{2} + x^{2} + x + x + 1 \cdot 1 = 25$

$y = x + 1$

Multiply $1$ by $1$ .

$x^{2} + x^{2} + x + x + 1 = 25$

$y = x + 1$

$x^{2} + x^{2} + x + x + 1 = 25$

$y = x + 1$

Add $x$ and $x$ .

$x^{2} + x^{2} + 2 x + 1 = 25$

$y = x + 1$

$x^{2} + x^{2} + 2 x + 1 = 25$

$y = x + 1$

$x^{2} + x^{2} + 2 x + 1 = 25$

$y = x + 1$

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

$2 x^{2} + 2 x + 1 = 25$

$y = x + 1$

$2 x^{2} + 2 x + 1 = 25$

$y = x + 1$

$2 x^{2} + 2 x + 1 = 25$

$y = x + 1$

$2 x^{2} + 2 x + 1 = 25$

$y = x + 1$

Step 2

Solve for $x$ in $2 x^{2} + 2 x + 1 = 25$ .

Subtract $25$ from both sides of the equation.

$2 x^{2} + 2 x + 1 - 25 = 0$

$y = x + 1$

Subtract $25$ from $1$ .

$2 x^{2} + 2 x - 24 = 0$

$y = x + 1$

Factor the left side of the equation.

Factor $2$ out of $2 x^{2} + 2 x - 24$ .

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

$2 (x^{2}) + 2 x - 24 = 0$

$y = x + 1$

Factor $2$ out of $2 x$ .

$2 (x^{2}) + 2 (x) - 24 = 0$

$y = x + 1$

Factor $2$ out of $- 24$ .

$2 (x^{2}) + 2 x + 2 \cdot - 12 = 0$

$y = x + 1$

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

$2 (x^{2} + x) + 2 \cdot - 12 = 0$

$y = x + 1$

Factor $2$ out of $2 (x^{2} + x) + 2 \cdot - 12$ .

$2 (x^{2} + x - 12) = 0$

$y = x + 1$

$2 (x^{2} + x - 12) = 0$

$y = x + 1$

Factor.

Factor $x^{2} + x - 12$ 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 $- 12$ and whose sum is $1$ .

$- 3, 4$

$y = x + 1$

Write the factored form using these integers.

$2 ((x - 3) (x + 4)) = 0$

$y = x + 1$

$2 ((x - 3) (x + 4)) = 0$

$y = x + 1$

Remove unnecessary parentheses.

$2 (x - 3) (x + 4) = 0$

$y = x + 1$

$2 (x - 3) (x + 4) = 0$

$y = x + 1$

$2 (x - 3) (x + 4) = 0$

$y = x + 1$

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

$x - 3 = 0$

$x + 4 = 0$

$y = x + 1$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

$y = x + 1$

Add $3$ to both sides of the equation.

$x = 3$

$y = x + 1$

$x = 3$

$y = x + 1$

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

$y = x + 1$

Subtract $4$ from both sides of the equation.

$x = - 4$

$y = x + 1$

$x = - 4$

$y = x + 1$

The final solution is all the values that make $2 (x - 3) (x + 4) = 0$ true.

$x = 3, - 4$

$y = x + 1$

$x = 3, - 4$

$y = x + 1$

Step 3

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

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

$y = (3) + 1$

$x = 3$

Simplify the right side.

Add $3$ and $1$ .

$y = 4$

$x = 3$

$y = 4$

$x = 3$

$y = 4$

$x = 3$

Step 4

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

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

$y = (- 4) + 1$

$x = - 4$

Simplify the right side.

Add $- 4$ and $1$ .

$y = - 3$

$x = - 4$

$y = - 3$

$x = - 4$

$y = - 3$

$x = - 4$

Step 5

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

$(3, 4)$

$(- 4, - 3)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(3, 4), (- 4, - 3)$

Equation Form:

$x = 3, y = 4$

$x = - 4, y = - 3$

Step 7

Solve by Substitution x^2+y^2=25 , y=x+1

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