Solve the System of Equations x^2+y^2=50 -2x+y=5

$x^{2} + y^{2} = 50$ $- 2 x + y = 5$

Step 1

Add $2 x$ to both sides of the equation.

$y = 5 + 2 x$

$x^{2} + y^{2} = 50$

Step 2

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

Replace all occurrences of $y$ in $x^{2} + y^{2} = 50$ with $5 + 2 x$ .

$x^{2} + {(5 + 2 x)}^{2} = 50$

$y = 5 + 2 x$

Simplify the left side.

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

Simplify each term.

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

$x^{2} + (5 + 2 x) (5 + 2 x) = 50$

$y = 5 + 2 x$

Expand $(5 + 2 x) (5 + 2 x)$ using the FOIL Method.

Apply the distributive property.

$x^{2} + 5 (5 + 2 x) + 2 x (5 + 2 x) = 50$

$y = 5 + 2 x$

Apply the distributive property.

$x^{2} + 5 \cdot 5 + 5 (2 x) + 2 x (5 + 2 x) = 50$

$y = 5 + 2 x$

Apply the distributive property.

$x^{2} + 5 \cdot 5 + 5 (2 x) + 2 x \cdot 5 + 2 x (2 x) = 50$

$y = 5 + 2 x$

$x^{2} + 5 \cdot 5 + 5 (2 x) + 2 x \cdot 5 + 2 x (2 x) = 50$

$y = 5 + 2 x$

Simplify and combine like terms.

Simplify each term.

Multiply $5$ by $5$ .

$x^{2} + 25 + 5 (2 x) + 2 x \cdot 5 + 2 x (2 x) = 50$

$y = 5 + 2 x$

Multiply $2$ by $5$ .

$x^{2} + 25 + 10 x + 2 x \cdot 5 + 2 x (2 x) = 50$

$y = 5 + 2 x$

Multiply $5$ by $2$ .

$x^{2} + 25 + 10 x + 10 x + 2 x (2 x) = 50$

$y = 5 + 2 x$

Rewrite using the commutative property of multiplication.

$x^{2} + 25 + 10 x + 10 x + 2 \cdot (2 x \cdot x) = 50$

$y = 5 + 2 x$

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

Move $x$ .

$x^{2} + 25 + 10 x + 10 x + 2 \cdot (2 (x \cdot x)) = 50$

$y = 5 + 2 x$

Multiply $x$ by $x$ .

$x^{2} + 25 + 10 x + 10 x + 2 \cdot (2 x^{2}) = 50$

$y = 5 + 2 x$

$x^{2} + 25 + 10 x + 10 x + 2 \cdot (2 x^{2}) = 50$

$y = 5 + 2 x$

Multiply $2$ by $2$ .

$x^{2} + 25 + 10 x + 10 x + 4 x^{2} = 50$

$y = 5 + 2 x$

$x^{2} + 25 + 10 x + 10 x + 4 x^{2} = 50$

$y = 5 + 2 x$

Add $10 x$ and $10 x$ .

$x^{2} + 25 + 20 x + 4 x^{2} = 50$

$y = 5 + 2 x$

$x^{2} + 25 + 20 x + 4 x^{2} = 50$

$y = 5 + 2 x$

$x^{2} + 25 + 20 x + 4 x^{2} = 50$

$y = 5 + 2 x$

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

$5 x^{2} + 25 + 20 x = 50$

$y = 5 + 2 x$

$5 x^{2} + 25 + 20 x = 50$

$y = 5 + 2 x$

$5 x^{2} + 25 + 20 x = 50$

$y = 5 + 2 x$

$5 x^{2} + 25 + 20 x = 50$

$y = 5 + 2 x$

Step 3

Solve for $x$ in $5 x^{2} + 25 + 20 x = 50$ .

Subtract $50$ from both sides of the equation.

$5 x^{2} + 25 + 20 x - 50 = 0$

$y = 5 + 2 x$

Subtract $50$ from $25$ .

$5 x^{2} + 20 x - 25 = 0$

$y = 5 + 2 x$

Factor the left side of the equation.

Factor $5$ out of $5 x^{2} + 20 x - 25$ .

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

$5 (x^{2}) + 20 x - 25 = 0$

$y = 5 + 2 x$

Factor $5$ out of $20 x$ .

$5 (x^{2}) + 5 (4 x) - 25 = 0$

$y = 5 + 2 x$

Factor $5$ out of $- 25$ .

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

$y = 5 + 2 x$

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

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

$y = 5 + 2 x$

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

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

$y = 5 + 2 x$

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

$y = 5 + 2 x$

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 $4$ .

$- 1, 5$

$y = 5 + 2 x$

Write the factored form using these integers.

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

$y = 5 + 2 x$

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

$y = 5 + 2 x$

Remove unnecessary parentheses.

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

$y = 5 + 2 x$

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

$y = 5 + 2 x$

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

$y = 5 + 2 x$

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

$x - 1 = 0$

$x + 5 = 0$

$y = 5 + 2 x$

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

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

$x - 1 = 0$

$y = 5 + 2 x$

Add $1$ to both sides of the equation.

$x = 1$

$y = 5 + 2 x$

$x = 1$

$y = 5 + 2 x$

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

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

$x + 5 = 0$

$y = 5 + 2 x$

Subtract $5$ from both sides of the equation.

$x = - 5$

$y = 5 + 2 x$

$x = - 5$

$y = 5 + 2 x$

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

$x = 1, - 5$

$y = 5 + 2 x$

$x = 1, - 5$

$y = 5 + 2 x$

Step 4

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

Replace all occurrences of $x$ in $y = 5 + 2 x$ with $1$ .

$y = 5 + 2 (1)$

$x = 1$

Simplify the right side.

Simplify $5 + 2 (1)$ .

Multiply $2$ by $1$ .

$y = 5 + 2$

$x = 1$

Add $5$ and $2$ .

$y = 7$

$x = 1$

$y = 7$

$x = 1$

$y = 7$

$x = 1$

$y = 7$

$x = 1$

Step 5

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

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

$y = 5 + 2 (- 5)$

$x = - 5$

Simplify the right side.

Simplify $5 + 2 (- 5)$ .

Multiply $2$ by $- 5$ .

$y = 5 - 10$

$x = - 5$

Subtract $10$ from $5$ .

$y = - 5$

$x = - 5$

$y = - 5$

$x = - 5$

$y = - 5$

$x = - 5$

$y = - 5$

$x = - 5$

Step 6

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

$(1, 7)$

$(- 5, - 5)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(1, 7), (- 5, - 5)$

Equation Form:

$x = 1, y = 7$

$x = - 5, y = - 5$

Step 8

Solve the System of Equations x^2+y^2=50 -2x+y=5

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