Find the Points of Intersection y=x+16 x^2+y^2=128

$y = x + 16$ $x^{2} + y^{2} = 128$

Step 1

Replace all occurrences of $y$ with $x + 16$ in each equation.

Replace all occurrences of $y$ in $x^{2} + y^{2} = 128$ with $x + 16$ .

$x^{2} + {(x + 16)}^{2} = 128$

$y = x + 16$

Simplify the left side.

Simplify $x^{2} + {(x + 16)}^{2}$ .

Simplify each term.

Rewrite ${(x + 16)}^{2}$ as $(x + 16) (x + 16)$ .

$x^{2} + (x + 16) (x + 16) = 128$

$y = x + 16$

Expand $(x + 16) (x + 16)$ using the FOIL Method.

Apply the distributive property.

$x^{2} + x (x + 16) + 16 (x + 16) = 128$

$y = x + 16$

Apply the distributive property.

$x^{2} + x \cdot x + x \cdot 16 + 16 (x + 16) = 128$

$y = x + 16$

Apply the distributive property.

$x^{2} + x \cdot x + x \cdot 16 + 16 x + 16 \cdot 16 = 128$

$y = x + 16$

$x^{2} + x \cdot x + x \cdot 16 + 16 x + 16 \cdot 16 = 128$

$y = x + 16$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + x^{2} + x \cdot 16 + 16 x + 16 \cdot 16 = 128$

$y = x + 16$

Move $16$ to the left of $x$ .

$x^{2} + x^{2} + 16 \cdot x + 16 x + 16 \cdot 16 = 128$

$y = x + 16$

Multiply $16$ by $16$ .

$x^{2} + x^{2} + 16 x + 16 x + 256 = 128$

$y = x + 16$

$x^{2} + x^{2} + 16 x + 16 x + 256 = 128$

$y = x + 16$

Add $16 x$ and $16 x$ .

$x^{2} + x^{2} + 32 x + 256 = 128$

$y = x + 16$

$x^{2} + x^{2} + 32 x + 256 = 128$

$y = x + 16$

$x^{2} + x^{2} + 32 x + 256 = 128$

$y = x + 16$

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

$2 x^{2} + 32 x + 256 = 128$

$y = x + 16$

$2 x^{2} + 32 x + 256 = 128$

$y = x + 16$

$2 x^{2} + 32 x + 256 = 128$

$y = x + 16$

$2 x^{2} + 32 x + 256 = 128$

$y = x + 16$

Step 2

Solve for $x$ in $2 x^{2} + 32 x + 256 = 128$ .

Subtract $128$ from both sides of the equation.

$2 x^{2} + 32 x + 256 - 128 = 0$

$y = x + 16$

Subtract $128$ from $256$ .

$2 x^{2} + 32 x + 128 = 0$

$y = x + 16$

Factor the left side of the equation.

Factor $2$ out of $2 x^{2} + 32 x + 128$ .

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

$2 (x^{2}) + 32 x + 128 = 0$

$y = x + 16$

Factor $2$ out of $32 x$ .

$2 (x^{2}) + 2 (16 x) + 128 = 0$

$y = x + 16$

Factor $2$ out of $128$ .

$2 x^{2} + 2 (16 x) + 2 \cdot 64 = 0$

$y = x + 16$

Factor $2$ out of $2 x^{2} + 2 (16 x)$ .

$2 (x^{2} + 16 x) + 2 \cdot 64 = 0$

$y = x + 16$

Factor $2$ out of $2 (x^{2} + 16 x) + 2 \cdot 64$ .

$2 (x^{2} + 16 x + 64) = 0$

$y = x + 16$

$2 (x^{2} + 16 x + 64) = 0$

$y = x + 16$

Factor using the perfect square rule.

Rewrite $64$ as $8^{2}$ .

$2 (x^{2} + 16 x + 8^{2}) = 0$

$y = x + 16$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$16 x = 2 \cdot x \cdot 8$

$y = x + 16$

Rewrite the polynomial.

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

$y = x + 16$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 8$ .

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

$y = x + 16$

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

$y = x + 16$

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

$y = x + 16$

Divide each term in $2 {(x + 8)}^{2} = 0$ by $2$ and simplify.

Divide each term in $2 {(x + 8)}^{2} = 0$ by $2$ .

$\frac{2 {(x + 8)}^{2}}{2} = \frac{0}{2}$

$y = x + 16$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 {(x + 8)}^{2}}{2} = \frac{0}{2}$

$y = x + 16$

Divide ${(x + 8)}^{2}$ by $1$ .

${(x + 8)}^{2} = \frac{0}{2}$

$y = x + 16$

${(x + 8)}^{2} = \frac{0}{2}$

$y = x + 16$

${(x + 8)}^{2} = \frac{0}{2}$

$y = x + 16$

Simplify the right side.

Divide $0$ by $2$ .

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

$y = x + 16$

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

$y = x + 16$

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

$y = x + 16$

Set the $x + 8$ equal to $0$ .

$x + 8 = 0$

$y = x + 16$

Subtract $8$ from both sides of the equation.

$x = - 8$

$y = x + 16$

$x = - 8$

$y = x + 16$

Step 3

Replace all occurrences of $x$ with $- 8$ in each equation.

Replace all occurrences of $x$ in $y = x + 16$ with $- 8$ .

$y = (- 8) + 16$

$x = - 8$

Simplify the right side.

Add $- 8$ and $16$ .

$y = 8$

$x = - 8$

$y = 8$

$x = - 8$

$y = 8$

$x = - 8$

Step 4

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

$(- 8, 8)$

Step 5

The result can be shown in multiple forms.

Point Form:

$(- 8, 8)$

Equation Form:

$x = - 8, y = 8$

Step 6

Find the Points of Intersection y=x+16 x^2+y^2=128

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