Precalculus Examples

$y = x^{2}$ , $y - 9 = 0$

Step 1

Add $9$ to both sides of the equation.

$y = 9$

$y = x^{2}$

Step 2

Replace all occurrences of $y$ with $9$ in each equation.

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Step 2.1

Replace all occurrences of $y$ in $y = x^{2}$ with $9$ .

$9 = x^{2}$

$y = 9$

Step 2.2

Simplify the left side.

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Step 2.2.1

Remove parentheses.

$9 = x^{2}$

$y = 9$

$9 = x^{2}$

$y = 9$

$9 = x^{2}$

$y = 9$

Step 3

Solve for $x$ in $9 = x^{2}$ .

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Step 3.1

Rewrite the equation as $x^{2} = 9$ .

$x^{2} = 9$

$y = 9$

Step 3.2

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

$x = \pm \sqrt{9}$

$y = 9$

Step 3.3

Simplify $\pm \sqrt{9}$ .

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Step 3.3.1

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

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

$y = 9$

Step 3.3.2

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

$x = \pm 3$

$y = 9$

$x = \pm 3$

$y = 9$

Step 3.4

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

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Step 3.4.1

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

$x = 3$

$y = 9$

Step 3.4.2

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

$x = - 3$

$y = 9$

Step 3.4.3

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

$x = 3, - 3$

$y = 9$

$x = 3, - 3$

$y = 9$

$x = 3, - 3$

$y = 9$

Step 4

Solve the system of equations.

$x = 3, y = 9$

Step 5

Solve the system of equations.

$x = - 3, y = 9$

Step 6

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

$(3, 9)$

$(- 3, 9)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(3, 9), (- 3, 9)$

Equation Form:

$x = 3, y = 9$

$x = - 3, y = 9$

Step 8