Solve the System of Equations x^2+y^2=25 2x+y=10

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

Subtract $2 x$ from both sides of the equation.

$y = 10 - 2 x$

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

Replace all occurrences of $y$ with $10 - 2 x$ in each equation.

Replace all occurrences of $y$ in $x^{2} + y^{2} = 25$ with $10 - 2 x$ .

$x^{2} + {(10 - 2 x)}^{2} = 25$

$y = 10 - 2 x$

Simplify the left side.

Simplify $x^{2} + {(10 - 2 x)}^{2}$ .

Simplify each term.

Rewrite ${(10 - 2 x)}^{2}$ as $(10 - 2 x) (10 - 2 x)$ .

$x^{2} + (10 - 2 x) (10 - 2 x) = 25$

$y = 10 - 2 x$

Expand $(10 - 2 x) (10 - 2 x)$ using the FOIL Method.

Apply the distributive property.

$x^{2} + 10 (10 - 2 x) - 2 x (10 - 2 x) = 25$

$y = 10 - 2 x$

Apply the distributive property.

$x^{2} + 10 \cdot 10 + 10 (- 2 x) - 2 x (10 - 2 x) = 25$

$y = 10 - 2 x$

Apply the distributive property.

$x^{2} + 10 \cdot 10 + 10 (- 2 x) - 2 x \cdot 10 - 2 x (- 2 x) = 25$

$y = 10 - 2 x$

$x^{2} + 10 \cdot 10 + 10 (- 2 x) - 2 x \cdot 10 - 2 x (- 2 x) = 25$

$y = 10 - 2 x$

Simplify and combine like terms.

Simplify each term.

Multiply $10$ by $10$ .

$x^{2} + 100 + 10 (- 2 x) - 2 x \cdot 10 - 2 x (- 2 x) = 25$

$y = 10 - 2 x$

Multiply $- 2$ by $10$ .

$x^{2} + 100 - 20 x - 2 x \cdot 10 - 2 x (- 2 x) = 25$

$y = 10 - 2 x$

Multiply $10$ by $- 2$ .

$x^{2} + 100 - 20 x - 20 x - 2 x (- 2 x) = 25$

$y = 10 - 2 x$

Rewrite using the commutative property of multiplication.

$x^{2} + 100 - 20 x - 20 x - 2 \cdot (- 2 x \cdot x) = 25$

$y = 10 - 2 x$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$x^{2} + 100 - 20 x - 20 x - 2 \cdot (- 2 (x \cdot x)) = 25$

$y = 10 - 2 x$

Multiply $x$ by $x$ .

$x^{2} + 100 - 20 x - 20 x - 2 \cdot (- 2 x^{2}) = 25$

$y = 10 - 2 x$

$x^{2} + 100 - 20 x - 20 x - 2 \cdot (- 2 x^{2}) = 25$

$y = 10 - 2 x$

Multiply $- 2$ by $- 2$ .

$x^{2} + 100 - 20 x - 20 x + 4 x^{2} = 25$

$y = 10 - 2 x$

$x^{2} + 100 - 20 x - 20 x + 4 x^{2} = 25$

$y = 10 - 2 x$

Subtract $20 x$ from $- 20 x$ .

$x^{2} + 100 - 40 x + 4 x^{2} = 25$

$y = 10 - 2 x$

$x^{2} + 100 - 40 x + 4 x^{2} = 25$

$y = 10 - 2 x$

$x^{2} + 100 - 40 x + 4 x^{2} = 25$

$y = 10 - 2 x$

Add $x^{2}$ and $4 x^{2}$ .

$5 x^{2} + 100 - 40 x = 25$

$y = 10 - 2 x$

$5 x^{2} + 100 - 40 x = 25$

$y = 10 - 2 x$

$5 x^{2} + 100 - 40 x = 25$

$y = 10 - 2 x$

$5 x^{2} + 100 - 40 x = 25$

$y = 10 - 2 x$

Solve for $x$ in $5 x^{2} + 100 - 40 x = 25$ .

Subtract $25$ from both sides of the equation.

$5 x^{2} + 100 - 40 x - 25 = 0$

$y = 10 - 2 x$

Subtract $25$ from $100$ .

$5 x^{2} - 40 x + 75 = 0$

$y = 10 - 2 x$

Factor the left side of the equation.

Factor $5$ out of $5 x^{2} - 40 x + 75$ .

Factor $5$ out of $5 x^{2}$ .

$5 (x^{2}) - 40 x + 75 = 0$

$y = 10 - 2 x$

Factor $5$ out of $- 40 x$ .

$5 (x^{2}) + 5 (- 8 x) + 75 = 0$

$y = 10 - 2 x$

Factor $5$ out of $75$ .

$5 x^{2} + 5 (- 8 x) + 5 \cdot 15 = 0$

$y = 10 - 2 x$

Factor $5$ out of $5 x^{2} + 5 (- 8 x)$ .

$5 (x^{2} - 8 x) + 5 \cdot 15 = 0$

$y = 10 - 2 x$

Factor $5$ out of $5 (x^{2} - 8 x) + 5 \cdot 15$ .

$5 (x^{2} - 8 x + 15) = 0$

$y = 10 - 2 x$

$5 (x^{2} - 8 x + 15) = 0$

$y = 10 - 2 x$

Factor.

Factor $x^{2} - 8 x + 15$ 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 $15$ and whose sum is $- 8$ .

$- 5, - 3$

$y = 10 - 2 x$

Write the factored form using these integers.

$5 ((x - 5) (x - 3)) = 0$

$y = 10 - 2 x$

$5 ((x - 5) (x - 3)) = 0$

$y = 10 - 2 x$

Remove unnecessary parentheses.

$5 (x - 5) (x - 3) = 0$

$y = 10 - 2 x$

$5 (x - 5) (x - 3) = 0$

$y = 10 - 2 x$

$5 (x - 5) (x - 3) = 0$

$y = 10 - 2 x$

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

$x - 5 = 0$

$x - 3 = 0$

$y = 10 - 2 x$

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

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

$x - 5 = 0$

$y = 10 - 2 x$

Add $5$ to both sides of the equation.

$x = 5$

$y = 10 - 2 x$

$x = 5$

$y = 10 - 2 x$

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

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

$x - 3 = 0$

$y = 10 - 2 x$

Add $3$ to both sides of the equation.

$x = 3$

$y = 10 - 2 x$

$x = 3$

$y = 10 - 2 x$

The final solution is all the values that make $5 (x - 5) (x - 3) = 0$ true.

$x = 5, 3$

$y = 10 - 2 x$

$x = 5, 3$

$y = 10 - 2 x$

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

Replace all occurrences of $x$ in $y = 10 - 2 x$ with $5$ .

$y = 10 - 2 \cdot 5$

$x = 5$

Simplify the right side.

Simplify $10 - 2 \cdot 5$ .

Multiply $- 2$ by $5$ .

$y = 10 - 10$

$x = 5$

Subtract $10$ from $10$ .

$y = 0$

$x = 5$

$y = 0$

$x = 5$

$y = 0$

$x = 5$

$y = 0$

$x = 5$

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

Replace all occurrences of $x$ in $y = 10 - 2 x$ with $3$ .

$y = 10 - 2 \cdot 3$

$x = 3$

Simplify the right side.

Simplify $10 - 2 \cdot 3$ .

Multiply $- 2$ by $3$ .

$y = 10 - 6$

$x = 3$

Subtract $6$ from $10$ .

$y = 4$

$x = 3$

$y = 4$

$x = 3$

$y = 4$

$x = 3$

$y = 4$

$x = 3$

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

$(5, 0)$

$(3, 4)$

The result can be shown in multiple forms.

Point Form:

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

Equation Form:

$x = 5, y = 0$

$x = 3, y = 4$

Solve the System of Equations x^2+y^2=25 2x+y=10

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