Algebra Examples

What is the solution to the system of equations $y = - x^{2} + 2$ and $y = x^{2}$ ?

Step 1

Eliminate the equal sides of each equation and combine.

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

Step 2

Solve $- x^{2} + 2 = x^{2}$ for $x$ .

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Step 2.1

Move all terms containing $x$ to the left side of the equation.

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Step 2.1.1

Subtract $x^{2}$ from both sides of the equation.

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

Step 2.1.2

Subtract $x^{2}$ from $- x^{2}$ .

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

Step 2.2

Subtract $2$ from both sides of the equation.

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

Step 2.3

Divide each term in $- 2 x^{2} = - 2$ by $- 2$ and simplify.

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Step 2.3.1

Divide each term in $- 2 x^{2} = - 2$ by $- 2$ .

$\frac{- 2 x^{2}}{- 2} = \frac{- 2}{- 2}$

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Cancel the common factor of $- 2$ .

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Step 2.3.2.1.1

Cancel the common factor.

$\frac{- 2 x^{2}}{- 2} = \frac{- 2}{- 2}$

Step 2.3.2.1.2

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{- 2}{- 2}$

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Divide $- 2$ by $- 2$ .

$x^{2} = 1$

Step 2.4

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{1}$

Step 2.5

Any root of $1$ is $1$ .

$x = \pm 1$

Step 2.6

The complete solution is the result of both the positive and negative portions of the solution.

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Step 2.6.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 1$

Step 2.6.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 1$

Step 2.6.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 1, - 1$

Step 3

Evaluate $y$ when $x = 1$ .

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Step 3.1

Substitute $1$ for $x$ .

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

Step 3.2

Substitute $1$ for $x$ in $y = {(1)}^{2}$ and solve for $y$ .

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Step 3.2.1

Remove parentheses.

$y = 1^{2}$

Step 3.2.2

One to any power is one.

$y = 1$

Step 4

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

$(1, 1)$

$(- 1, 1)$

Step 5

The result can be shown in multiple forms.

Point Form:

$(1, 1), (- 1, 1)$

Equation Form:

$x = 1, y = 1$

$x = - 1, y = 1$

Step 6