Precalculus Examples

$y = \frac{- 32 x^{2}}{160^{2}} + x$ , $y^{2} = 100$

Step 1

Simplify each term.

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Raise $160$ to the power of $2$ .

$y = \frac{- 32 x^{2}}{25600} + x$

$y^{2} = 100$

Cancel the common factor of $- 32$ and $25600$ .

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Factor $32$ out of $- 32 x^{2}$ .

$y = \frac{32 (- x^{2})}{25600} + x$

$y^{2} = 100$

Cancel the common factors.

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Factor $32$ out of $25600$ .

$y = \frac{32 (- x^{2})}{32 (800)} + x$

$y^{2} = 100$

Cancel the common factor.

$y = \frac{32 (- x^{2})}{32 \cdot 800} + x$

$y^{2} = 100$

Rewrite the expression.

$y = \frac{- x^{2}}{800} + x$

$y^{2} = 100$

$y = \frac{- x^{2}}{800} + x$

$y^{2} = 100$

$y = \frac{- x^{2}}{800} + x$

$y^{2} = 100$

Move the negative in front of the fraction.

$y = - \frac{x^{2}}{800} + x$

$y^{2} = 100$

$y = - \frac{x^{2}}{800} + x$

$y^{2} = 100$

Step 2

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

$y = \pm \sqrt{100}$

$y = - \frac{x^{2}}{800} + x$

Step 3

Simplify $\pm \sqrt{100}$ .

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Rewrite $100$ as $10^{2}$ .

$y = \pm \sqrt{10^{2}}$

$y = - \frac{x^{2}}{800} + x$

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

$y = \pm 10$

$y = - \frac{x^{2}}{800} + x$

$y = \pm 10$

$y = - \frac{x^{2}}{800} + x$

Step 4

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

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

$y = 10$

$y = - \frac{x^{2}}{800} + x$

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

$y = - 10$

$y = - \frac{x^{2}}{800} + x$

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

$y = 10, - 10$

$y = - \frac{x^{2}}{800} + x$

$y = 10, - 10$

$y = - \frac{x^{2}}{800} + x$

Step 5

Create a graph to locate the intersection of the equations. The intersection of the system of equations is the solution.

$(400 + 40 \sqrt{95}, 10)$

$(400 + 40 \sqrt{105}, - 10)$

Step 6