Precalculus Examples

$x y = 12$ , $x^{2} + y^{2} = 40$

Step 1

Divide each term in $x y = 12$ by $y$ and simplify.

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Step 1.1

Divide each term in $x y = 12$ by $y$ .

$\frac{x y}{y} = \frac{12}{y}$

$x^{2} + y^{2} = 40$

Step 1.2

Simplify the left side.

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Step 1.2.1

Cancel the common factor of $y$ .

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Step 1.2.1.1

Cancel the common factor.

$\frac{x y}{y} = \frac{12}{y}$

$x^{2} + y^{2} = 40$

Step 1.2.1.2

Divide $x$ by $1$ .

$x = \frac{12}{y}$

$x^{2} + y^{2} = 40$

$x = \frac{12}{y}$

$x^{2} + y^{2} = 40$

$x = \frac{12}{y}$

$x^{2} + y^{2} = 40$

$x = \frac{12}{y}$

$x^{2} + y^{2} = 40$

Step 2

Replace all occurrences of $x$ with $\frac{12}{y}$ in each equation.

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Step 2.1

Replace all occurrences of $x$ in $x^{2} + y^{2} = 40$ with $\frac{12}{y}$ .

${(\frac{12}{y})}^{2} + y^{2} = 40$

$x = \frac{12}{y}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify each term.

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Step 2.2.1.1

Apply the product rule to $\frac{12}{y}$ .

$\frac{12^{2}}{y^{2}} + y^{2} = 40$

$x = \frac{12}{y}$

Step 2.2.1.2

Raise $12$ to the power of $2$ .

$\frac{144}{y^{2}} + y^{2} = 40$

$x = \frac{12}{y}$

$\frac{144}{y^{2}} + y^{2} = 40$

$x = \frac{12}{y}$

$\frac{144}{y^{2}} + y^{2} = 40$

$x = \frac{12}{y}$

$\frac{144}{y^{2}} + y^{2} = 40$

$x = \frac{12}{y}$

Step 3

Solve for $y$ in $\frac{144}{y^{2}} + y^{2} = 40$ .

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Step 3.1

Find the LCD of the terms in the equation.

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Step 3.1.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$y^{2}, 1, 1$

$x = \frac{12}{y}$

Step 3.1.2

The LCM of one and any expression is the expression.

$y^{2}$

$x = \frac{12}{y}$

$y^{2}$

$x = \frac{12}{y}$

Step 3.2

Multiply each term in $\frac{144}{y^{2}} + y^{2} = 40$ by $y^{2}$ to eliminate the fractions.

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Step 3.2.1

Multiply each term in $\frac{144}{y^{2}} + y^{2} = 40$ by $y^{2}$ .

$\frac{144}{y^{2}} \cdot y^{2} + y^{2} y^{2} = 40 y^{2}$

$x = \frac{12}{y}$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Simplify each term.

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Step 3.2.2.1.1

Cancel the common factor of $y^{2}$ .

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Step 3.2.2.1.1.1

Cancel the common factor.

$\frac{144}{y^{2}} \cdot y^{2} + y^{2} y^{2} = 40 y^{2}$

$x = \frac{12}{y}$

Step 3.2.2.1.1.2

Rewrite the expression.

$144 + y^{2} y^{2} = 40 y^{2}$

$x = \frac{12}{y}$

$144 + y^{2} y^{2} = 40 y^{2}$

$x = \frac{12}{y}$

Step 3.2.2.1.2

Multiply $y^{2}$ by $y^{2}$ by adding the exponents.

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Step 3.2.2.1.2.1

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$144 + y^{2 + 2} = 40 y^{2}$

$x = \frac{12}{y}$

Step 3.2.2.1.2.2

Add $2$ and $2$ .

$144 + y^{4} = 40 y^{2}$

$x = \frac{12}{y}$

$144 + y^{4} = 40 y^{2}$

$x = \frac{12}{y}$

$144 + y^{4} = 40 y^{2}$

$x = \frac{12}{y}$

$144 + y^{4} = 40 y^{2}$

$x = \frac{12}{y}$

$144 + y^{4} = 40 y^{2}$

$x = \frac{12}{y}$

Step 3.3

Solve the equation.

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Step 3.3.1

Subtract $40 y^{2}$ from both sides of the equation.

$144 + y^{4} - 40 y^{2} = 0$

$x = \frac{12}{y}$

Step 3.3.2

Substitute $u = y^{2}$ into the equation. This will make the quadratic formula easy to use.

$u^{2} - 40 u + 144 = 0$

$u = y^{2}$

$x = \frac{12}{y}$

Step 3.3.3

Factor $u^{2} - 40 u + 144$ using the AC method.

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Step 3.3.3.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $144$ and whose sum is $- 40$ .

$- 36, - 4$

$x = \frac{12}{y}$

Step 3.3.3.2

Write the factored form using these integers.

$(u - 36) (u - 4) = 0$

$x = \frac{12}{y}$

$(u - 36) (u - 4) = 0$

$x = \frac{12}{y}$

Step 3.3.4

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

$u - 36 = 0$

$u - 4 = 0$

$x = \frac{12}{y}$

Step 3.3.5

Set $u - 36$ equal to $0$ and solve for $u$ .

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Step 3.3.5.1

Set $u - 36$ equal to $0$ .

$u - 36 = 0$

$x = \frac{12}{y}$

Step 3.3.5.2

Add $36$ to both sides of the equation.

$u = 36$

$x = \frac{12}{y}$

$u = 36$

$x = \frac{12}{y}$

Step 3.3.6

Set $u - 4$ equal to $0$ and solve for $u$ .

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Step 3.3.6.1

Set $u - 4$ equal to $0$ .

$u - 4 = 0$

$x = \frac{12}{y}$

Step 3.3.6.2

Add $4$ to both sides of the equation.

$u = 4$

$x = \frac{12}{y}$

$u = 4$

$x = \frac{12}{y}$

Step 3.3.7

The final solution is all the values that make $(u - 36) (u - 4) = 0$ true.

$u = 36, 4$

$x = \frac{12}{y}$

Step 3.3.8

Substitute the real value of $u = y^{2}$ back into the solved equation.

$y^{2} = 36$

$y^{2} = 4$

$x = \frac{12}{y}$

Step 3.3.9

Solve the first equation for $y$ .

$y^{2} = 36$

$x = \frac{12}{y}$

Step 3.3.10

Solve the equation for $y$ .

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Step 3.3.10.1

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

$y = \pm \sqrt{36}$

$x = \frac{12}{y}$

Step 3.3.10.2

Simplify $\pm \sqrt{36}$ .

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Step 3.3.10.2.1

Rewrite $36$ as $6^{2}$ .

$y = \pm \sqrt{6^{2}}$

$x = \frac{12}{y}$

Step 3.3.10.2.2

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

$y = \pm 6$

$x = \frac{12}{y}$

$y = \pm 6$

$x = \frac{12}{y}$

Step 3.3.10.3

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

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Step 3.3.10.3.1

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

$y = 6$

$x = \frac{12}{y}$

Step 3.3.10.3.2

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

$y = - 6$

$x = \frac{12}{y}$

Step 3.3.10.3.3

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

$y = 6, - 6$

$x = \frac{12}{y}$

$y = 6, - 6$

$x = \frac{12}{y}$

$y = 6, - 6$

$x = \frac{12}{y}$

Step 3.3.11

Solve the second equation for $y$ .

$y^{2} = 4$

$x = \frac{12}{y}$

Step 3.3.12

Solve the equation for $y$ .

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Step 3.3.12.1

Remove parentheses.

$y^{2} = 4$

$x = \frac{12}{y}$

Step 3.3.12.2

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

$y = \pm \sqrt{4}$

$x = \frac{12}{y}$

Step 3.3.12.3

Simplify $\pm \sqrt{4}$ .

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Step 3.3.12.3.1

Rewrite $4$ as $2^{2}$ .

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

$x = \frac{12}{y}$

Step 3.3.12.3.2

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

$y = \pm 2$

$x = \frac{12}{y}$

$y = \pm 2$

$x = \frac{12}{y}$

Step 3.3.12.4

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

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Step 3.3.12.4.1

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

$y = 2$

$x = \frac{12}{y}$

Step 3.3.12.4.2

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

$y = - 2$

$x = \frac{12}{y}$

Step 3.3.12.4.3

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

$y = 2, - 2$

$x = \frac{12}{y}$

$y = 2, - 2$

$x = \frac{12}{y}$

$y = 2, - 2$

$x = \frac{12}{y}$

Step 3.3.13

The solution to $y^{4} - 40 y^{2} + 144 = 0$ is $y = 6, - 6, 2, - 2$ .

$y = 6, - 6, 2, - 2$

$x = \frac{12}{y}$

$y = 6, - 6, 2, - 2$

$x = \frac{12}{y}$

$y = 6, - 6, 2, - 2$

$x = \frac{12}{y}$

Step 4

Replace all occurrences of $y$ with $6$ in each equation.

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Step 4.1

Replace all occurrences of $y$ in $x = \frac{12}{y}$ with $6$ .

$x = \frac{12}{6}$

$y = 6$

Step 4.2

Simplify the right side.

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Step 4.2.1

Divide $12$ by $6$ .

$x = 2$

$y = 6$

$x = 2$

$y = 6$

$x = 2$

$y = 6$

Step 5

Replace all occurrences of $y$ with $- 6$ in each equation.

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Step 5.1

Replace all occurrences of $y$ in $x = \frac{12}{y}$ with $- 6$ .

$x = \frac{12}{- 6}$

$y = - 6$

Step 5.2

Simplify the right side.

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Step 5.2.1

Divide $12$ by $- 6$ .

$x = - 2$

$y = - 6$

$x = - 2$

$y = - 6$

$x = - 2$

$y = - 6$

Step 6

Replace all occurrences of $y$ with $6$ in each equation.

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Step 6.1

Replace all occurrences of $y$ in $x = \frac{12}{y}$ with $6$ .

$x = \frac{12}{6}$

$y = 6$

Step 6.2

Simplify the right side.

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Step 6.2.1

Divide $12$ by $6$ .

$x = 2$

$y = 6$

$x = 2$

$y = 6$

$x = 2$

$y = 6$

Step 7

Replace all occurrences of $y$ with $- 6$ in each equation.

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Step 7.1

Replace all occurrences of $y$ in $x = \frac{12}{y}$ with $- 6$ .

$x = \frac{12}{- 6}$

$y = - 6$

Step 7.2

Simplify the right side.

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Step 7.2.1

Divide $12$ by $- 6$ .

$x = - 2$

$y = - 6$

$x = - 2$

$y = - 6$

$x = - 2$

$y = - 6$

Step 8

Replace all occurrences of $y$ with $2$ in each equation.

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Step 8.1

Replace all occurrences of $y$ in $x = \frac{12}{y}$ with $2$ .

$x = \frac{12}{2}$

$y = 2$

Step 8.2

Simplify the right side.

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Step 8.2.1

Divide $12$ by $2$ .

$x = 6$

$y = 2$

$x = 6$

$y = 2$

$x = 6$

$y = 2$

Step 9

Replace all occurrences of $y$ with $6$ in each equation.

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Step 9.1

Replace all occurrences of $y$ in $x = \frac{12}{y}$ with $6$ .

$x = \frac{12}{6}$

$y = 6$

Step 9.2

Simplify the right side.

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Step 9.2.1

Divide $12$ by $6$ .

$x = 2$

$y = 6$

$x = 2$

$y = 6$

$x = 2$

$y = 6$

Step 10

Replace all occurrences of $y$ with $- 6$ in each equation.

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Step 10.1

Replace all occurrences of $y$ in $x = \frac{12}{y}$ with $- 6$ .

$x = \frac{12}{- 6}$

$y = - 6$

Step 10.2

Simplify the right side.

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Step 10.2.1

Divide $12$ by $- 6$ .

$x = - 2$

$y = - 6$

$x = - 2$

$y = - 6$

$x = - 2$

$y = - 6$

Step 11

Replace all occurrences of $y$ with $2$ in each equation.

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Step 11.1

Replace all occurrences of $y$ in $x = \frac{12}{y}$ with $2$ .

$x = \frac{12}{2}$

$y = 2$

Step 11.2

Simplify the right side.

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Step 11.2.1

Divide $12$ by $2$ .

$x = 6$

$y = 2$

$x = 6$

$y = 2$

$x = 6$

$y = 2$

Step 12

Replace all occurrences of $y$ with $- 2$ in each equation.

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Step 12.1

Replace all occurrences of $y$ in $x = \frac{12}{y}$ with $- 2$ .

$x = \frac{12}{- 2}$

$y = - 2$

Step 12.2

Simplify the right side.

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Step 12.2.1

Divide $12$ by $- 2$ .

$x = - 6$

$y = - 2$

$x = - 6$

$y = - 2$

$x = - 6$

$y = - 2$

Step 13

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

$(2, 6)$

$(- 2, - 6)$

$(6, 2)$

$(- 6, - 2)$

Step 14

The result can be shown in multiple forms.

Point Form:

$(2, 6), (- 2, - 6), (6, 2), (- 6, - 2)$

Equation Form:

$x = 2, y = 6$

$x = - 2, y = - 6$

$x = 6, y = 2$

$x = - 6, y = - 2$

Step 15