Algebra Examples

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

Step 1

Eliminate the equal sides of each equation and combine.

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

Step 2

Solve $- {(x - 2)}^{2} + 4 = - 5$ for $x$ .

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Step 2.1

Subtract $4$ from both sides of the equation.

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

Step 2.2

Subtract $4$ from $- 5$ .

$- {(x - 2)}^{2} = - 9$

Step 2.3

Divide each term in $- {(x - 2)}^{2} = - 9$ by $- 1$ and simplify.

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Step 2.3.1

Divide each term in $- {(x - 2)}^{2} = - 9$ by $- 1$ .

$\frac{- {(x - 2)}^{2}}{- 1} = \frac{- 9}{- 1}$

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Dividing two negative values results in a positive value.

$\frac{{(x - 2)}^{2}}{1} = \frac{- 9}{- 1}$

Step 2.3.2.2

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

${(x - 2)}^{2} = \frac{- 9}{- 1}$

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Divide $- 9$ by $- 1$ .

${(x - 2)}^{2} = 9$

Step 2.4

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

$x - 2 = \pm \sqrt{9}$

Step 2.5

Simplify $\pm \sqrt{9}$ .

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Step 2.5.1

Rewrite $9$ as $3^{2}$ .

$x - 2 = \pm \sqrt{3^{2}}$

Step 2.5.2

Pull terms out from under the radical, assuming positive real numbers.

$x - 2 = \pm 3$

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.

$x - 2 = 3$

Step 2.6.2

Move all terms not containing $x$ to the right side of the equation.

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Step 2.6.2.1

Add $2$ to both sides of the equation.

$x = 3 + 2$

Step 2.6.2.2

Add $3$ and $2$ .

$x = 5$

Step 2.6.3

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

$x - 2 = - 3$

Step 2.6.4

Move all terms not containing $x$ to the right side of the equation.

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Step 2.6.4.1

Add $2$ to both sides of the equation.

$x = - 3 + 2$

Step 2.6.4.2

Add $- 3$ and $2$ .

$x = - 1$

Step 2.6.5

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

$x = 5, - 1$

Step 3

Substitute $5$ for $x$ .

$y = - 5$

Step 4

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

$(5, - 5)$

$(- 1, - 5)$

Step 5

The result can be shown in multiple forms.

Point Form:

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

Equation Form:

$x = 5, y = - 5$

$x = - 1, y = - 5$

Step 6