Solve the System of Equations y=x^2 2y+6=2(x+3)

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Algebra

$y = x^{2}$ $2 y + 6 = 2 (x + 3)$

Replace all occurrences of $y$ with $x^{2}$ in each equation.

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Replace all occurrences of $y$ in $2 y + 6 = 2 (x + 3)$ with $x^{2}$ .

$2 (x^{2}) + 6 = 2 (x + 3)$

$y = x^{2}$

Simplify $2 (x^{2}) + 6 = 2 (x + 3)$ .

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Simplify the left side.

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Multiply $2$ by $x^{2}$ .

$2 x^{2} + 6 = 2 (x + 3)$

$y = x^{2}$

$2 x^{2} + 6 = 2 (x + 3)$

$y = x^{2}$

Simplify the right side.

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Simplify $2 (x + 3)$ .

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Apply the distributive property.

$2 x^{2} + 6 = 2 x + 2 \cdot 3$

$y = x^{2}$

Multiply $2$ by $3$ .

$2 x^{2} + 6 = 2 x + 6$

$y = x^{2}$

$2 x^{2} + 6 = 2 x + 6$

$y = x^{2}$

$2 x^{2} + 6 = 2 x + 6$

$y = x^{2}$

$2 x^{2} + 6 = 2 x + 6$

$y = x^{2}$

$2 x^{2} + 6 = 2 x + 6$

$y = x^{2}$

Solve for $x$ in $2 x^{2} + 6 = 2 x + 6$ .

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

$2 x^{2} + 6 - 2 x = 6$

$y = x^{2}$

Subtract $6$ from both sides of the equation.

$2 x^{2} + 6 - 2 x - 6 = 0$

$y = x^{2}$

Combine the opposite terms in $2 x^{2} + 6 - 2 x - 6$ .

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Subtract $6$ from $6$ .

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

$y = x^{2}$

Add $2 x^{2} - 2 x$ and $0$ .

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

$y = x^{2}$

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

$y = x^{2}$

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

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

$2 x (x) - 2 x = 0$

$y = x^{2}$

Factor $2 x$ out of $- 2 x$ .

$2 x (x) + 2 x (- 1) = 0$

$y = x^{2}$

Factor $2 x$ out of $2 x (x) + 2 x (- 1)$ .

$2 x (x - 1) = 0$

$y = x^{2}$

$2 x (x - 1) = 0$

$y = x^{2}$

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

$x = 0$

$x - 1 = 0$

$y = x^{2}$

Set $x$ equal to $0$ .

$x = 0$

$y = x^{2}$

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

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

$x - 1 = 0$

$y = x^{2}$

Add $1$ to both sides of the equation.

$x = 1$

$y = x^{2}$

$x = 1$

$y = x^{2}$

The final solution is all the values that make $2 x (x - 1) = 0$ true.

$x = 0, 1$

$y = x^{2}$

$x = 0, 1$

$y = x^{2}$

Replace all occurrences of $x$ with $0$ in each equation.

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Replace all occurrences of $x$ in $y = x^{2}$ with $0$ .

$y = {(0)}^{2}$

$x = 0$

Simplify the right side.

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Raising $0$ to any positive power yields $0$ .

$y = 0$

$x = 0$

$y = 0$

$x = 0$

$y = 0$

$x = 0$

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

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

$y = {(1)}^{2}$

$x = 1$

Simplify the right side.

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One to any power is one.

$y = 1$

$x = 1$

$y = 1$

$x = 1$

$y = 1$

$x = 1$

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

$(0, 0)$

$(1, 1)$

The result can be shown in multiple forms.

Point Form:

$(0, 0), (1, 1)$

Equation Form:

$x = 0, y = 0$

$x = 1, y = 1$

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