Solve the System of Equations x^2+y^2=25 x-3y=15

$x^{2} + y^{2} = 25$ $x - 3 y = 15$

Step 1

Add $3 y$ to both sides of the equation.

$x = 15 + 3 y$

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

Step 2

Replace all occurrences of $x$ with $15 + 3 y$ in each equation.

Replace all occurrences of $x$ in $x^{2} + y^{2} = 25$ with $15 + 3 y$ .

${(15 + 3 y)}^{2} + y^{2} = 25$

$x = 15 + 3 y$

Simplify the left side.

Simplify ${(15 + 3 y)}^{2} + y^{2}$ .

Simplify each term.

Rewrite ${(15 + 3 y)}^{2}$ as $(15 + 3 y) (15 + 3 y)$ .

$(15 + 3 y) (15 + 3 y) + y^{2} = 25$

$x = 15 + 3 y$

Expand $(15 + 3 y) (15 + 3 y)$ using the FOIL Method.

Apply the distributive property.

$15 (15 + 3 y) + 3 y (15 + 3 y) + y^{2} = 25$

$x = 15 + 3 y$

Apply the distributive property.

$15 \cdot 15 + 15 (3 y) + 3 y (15 + 3 y) + y^{2} = 25$

$x = 15 + 3 y$

Apply the distributive property.

$15 \cdot 15 + 15 (3 y) + 3 y \cdot 15 + 3 y (3 y) + y^{2} = 25$

$x = 15 + 3 y$

$15 \cdot 15 + 15 (3 y) + 3 y \cdot 15 + 3 y (3 y) + y^{2} = 25$

$x = 15 + 3 y$

Simplify and combine like terms.

Simplify each term.

Multiply $15$ by $15$ .

$225 + 15 (3 y) + 3 y \cdot 15 + 3 y (3 y) + y^{2} = 25$

$x = 15 + 3 y$

Multiply $3$ by $15$ .

$225 + 45 y + 3 y \cdot 15 + 3 y (3 y) + y^{2} = 25$

$x = 15 + 3 y$

Multiply $15$ by $3$ .

$225 + 45 y + 45 y + 3 y (3 y) + y^{2} = 25$

$x = 15 + 3 y$

Rewrite using the commutative property of multiplication.

$225 + 45 y + 45 y + 3 \cdot (3 y \cdot y) + y^{2} = 25$

$x = 15 + 3 y$

Multiply $y$ by $y$ by adding the exponents.

Move $y$ .

$225 + 45 y + 45 y + 3 \cdot (3 (y \cdot y)) + y^{2} = 25$

$x = 15 + 3 y$

Multiply $y$ by $y$ .

$225 + 45 y + 45 y + 3 \cdot (3 y^{2}) + y^{2} = 25$

$x = 15 + 3 y$

$225 + 45 y + 45 y + 3 \cdot (3 y^{2}) + y^{2} = 25$

$x = 15 + 3 y$

Multiply $3$ by $3$ .

$225 + 45 y + 45 y + 9 y^{2} + y^{2} = 25$

$x = 15 + 3 y$

$225 + 45 y + 45 y + 9 y^{2} + y^{2} = 25$

$x = 15 + 3 y$

Add $45 y$ and $45 y$ .

$225 + 90 y + 9 y^{2} + y^{2} = 25$

$x = 15 + 3 y$

$225 + 90 y + 9 y^{2} + y^{2} = 25$

$x = 15 + 3 y$

$225 + 90 y + 9 y^{2} + y^{2} = 25$

$x = 15 + 3 y$

Add $9 y^{2}$ and $y^{2}$ .

$225 + 90 y + 10 y^{2} = 25$

$x = 15 + 3 y$

$225 + 90 y + 10 y^{2} = 25$

$x = 15 + 3 y$

$225 + 90 y + 10 y^{2} = 25$

$x = 15 + 3 y$

$225 + 90 y + 10 y^{2} = 25$

$x = 15 + 3 y$

Step 3

Solve for $y$ in $225 + 90 y + 10 y^{2} = 25$ .

Subtract $25$ from both sides of the equation.

$225 + 90 y + 10 y^{2} - 25 = 0$

$x = 15 + 3 y$

Subtract $25$ from $225$ .

$90 y + 10 y^{2} + 200 = 0$

$x = 15 + 3 y$

Factor the left side of the equation.

Factor $10$ out of $90 y + 10 y^{2} + 200$ .

Factor $10$ out of $90 y$ .

$10 (9 y) + 10 y^{2} + 200 = 0$

$x = 15 + 3 y$

Factor $10$ out of $10 y^{2}$ .

$10 (9 y) + 10 (y^{2}) + 200 = 0$

$x = 15 + 3 y$

Factor $10$ out of $200$ .

$10 (9 y) + 10 y^{2} + 10 \cdot 20 = 0$

$x = 15 + 3 y$

Factor $10$ out of $10 (9 y) + 10 y^{2}$ .

$10 (9 y + y^{2}) + 10 \cdot 20 = 0$

$x = 15 + 3 y$

Factor $10$ out of $10 (9 y + y^{2}) + 10 \cdot 20$ .

$10 (9 y + y^{2} + 20) = 0$

$x = 15 + 3 y$

$10 (9 y + y^{2} + 20) = 0$

$x = 15 + 3 y$

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

$10 (9 u + u^{2} + 20) = 0$

$x = 15 + 3 y$

Factor $9 u + u^{2} + 20$ 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 $20$ and whose sum is $9$ .

$4, 5$

$x = 15 + 3 y$

Write the factored form using these integers.

$10 ((u + 4) (u + 5)) = 0$

$x = 15 + 3 y$

$10 ((u + 4) (u + 5)) = 0$

$x = 15 + 3 y$

Factor.

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

$10 ((y + 4) (y + 5)) = 0$

$x = 15 + 3 y$

Remove unnecessary parentheses.

$10 (y + 4) (y + 5) = 0$

$x = 15 + 3 y$

$10 (y + 4) (y + 5) = 0$

$x = 15 + 3 y$

$10 (y + 4) (y + 5) = 0$

$x = 15 + 3 y$

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

$y + 4 = 0$

$y + 5 = 0$

$x = 15 + 3 y$

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

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

$y + 4 = 0$

$x = 15 + 3 y$

Subtract $4$ from both sides of the equation.

$y = - 4$

$x = 15 + 3 y$

$y = - 4$

$x = 15 + 3 y$

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

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

$y + 5 = 0$

$x = 15 + 3 y$

Subtract $5$ from both sides of the equation.

$y = - 5$

$x = 15 + 3 y$

$y = - 5$

$x = 15 + 3 y$

The final solution is all the values that make $10 (y + 4) (y + 5) = 0$ true.

$y = - 4, - 5$

$x = 15 + 3 y$

$y = - 4, - 5$

$x = 15 + 3 y$

Step 4

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

Replace all occurrences of $y$ in $x = 15 + 3 y$ with $- 4$ .

$x = 15 + 3 (- 4)$

$y = - 4$

Simplify the right side.

Simplify $15 + 3 (- 4)$ .

Multiply $3$ by $- 4$ .

$x = 15 - 12$

$y = - 4$

Subtract $12$ from $15$ .

$x = 3$

$y = - 4$

$x = 3$

$y = - 4$

$x = 3$

$y = - 4$

$x = 3$

$y = - 4$

Step 5

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

Replace all occurrences of $y$ in $x = 15 + 3 y$ with $- 5$ .

$x = 15 + 3 (- 5)$

$y = - 5$

Simplify the right side.

Simplify $15 + 3 (- 5)$ .

Multiply $3$ by $- 5$ .

$x = 15 - 15$

$y = - 5$

Subtract $15$ from $15$ .

$x = 0$

$y = - 5$

$x = 0$

$y = - 5$

$x = 0$

$y = - 5$

$x = 0$

$y = - 5$

Step 6

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

$(3, - 4)$

$(0, - 5)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(3, - 4), (0, - 5)$

Equation Form:

$x = 3, y = - 4$

$x = 0, y = - 5$

Step 8

Solve the System of Equations x^2+y^2=25 x-3y=15

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