Precalculus Examples

$x^{2} + y^{2} = 2$ , $x y = 1$

Step 1

Divide each term in $x y = 1$ by $y$ and simplify.

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Step 1.1

Divide each term in $x y = 1$ by $y$ .

$\frac{x y}{y} = \frac{1}{y}$

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

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.

$\frac{x y}{y} = \frac{1}{y}$

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

Step 1.2.1.2

Divide $x$ by $1$ .

$x = \frac{1}{y}$

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

$x = \frac{1}{y}$

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

$x = \frac{1}{y}$

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

$x = \frac{1}{y}$

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

Step 2

Replace all occurrences of $x$ with $\frac{1}{y}$ in each equation.

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Step 2.1

Replace all occurrences of $x$ in $x^{2} + y^{2} = 2$ with $\frac{1}{y}$ .

${(\frac{1}{y})}^{2} + y^{2} = 2$

$x = \frac{1}{y}$

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 $\frac{1}{y}$ .

$\frac{1^{2}}{y^{2}} + y^{2} = 2$

$x = \frac{1}{y}$

Step 2.2.1.2

One to any power is one.

$\frac{1}{y^{2}} + y^{2} = 2$

$x = \frac{1}{y}$

$\frac{1}{y^{2}} + y^{2} = 2$

$x = \frac{1}{y}$

$\frac{1}{y^{2}} + y^{2} = 2$

$x = \frac{1}{y}$

$\frac{1}{y^{2}} + y^{2} = 2$

$x = \frac{1}{y}$

Step 3

Solve for $y$ in $\frac{1}{y^{2}} + y^{2} = 2$ .

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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.

$y^{2}, 1, 1$

$x = \frac{1}{y}$

Step 3.1.2

The LCM of one and any expression is the expression.

$y^{2}$

$x = \frac{1}{y}$

$y^{2}$

$x = \frac{1}{y}$

Step 3.2

Multiply each term in $\frac{1}{y^{2}} + y^{2} = 2$ by $y^{2}$ to eliminate the fractions.

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Step 3.2.1

Multiply each term in $\frac{1}{y^{2}} + y^{2} = 2$ by $y^{2}$ .

$\frac{1}{y^{2}} \cdot y^{2} + y^{2} y^{2} = 2 y^{2}$

$x = \frac{1}{y}$

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.

$\frac{1}{y^{2}} \cdot y^{2} + y^{2} y^{2} = 2 y^{2}$

$x = \frac{1}{y}$

Step 3.2.2.1.1.2

Rewrite the expression.

$1 + y^{2} y^{2} = 2 y^{2}$

$x = \frac{1}{y}$

$1 + y^{2} y^{2} = 2 y^{2}$

$x = \frac{1}{y}$

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.

$1 + y^{2 + 2} = 2 y^{2}$

$x = \frac{1}{y}$

Step 3.2.2.1.2.2

Add $2$ and $2$ .

$1 + y^{4} = 2 y^{2}$

$x = \frac{1}{y}$

$1 + y^{4} = 2 y^{2}$

$x = \frac{1}{y}$

$1 + y^{4} = 2 y^{2}$

$x = \frac{1}{y}$

$1 + y^{4} = 2 y^{2}$

$x = \frac{1}{y}$

$1 + y^{4} = 2 y^{2}$

$x = \frac{1}{y}$

Step 3.3

Solve the equation.

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Step 3.3.1

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

$1 + y^{4} - 2 y^{2} = 0$

$x = \frac{1}{y}$

Step 3.3.2

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

$u^{2} - 2 u + 1 = 0$

$u = y^{2}$

$x = \frac{1}{y}$

Step 3.3.3

Factor using the perfect square rule.

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Step 3.3.3.1

Rewrite $1$ as $1^{2}$ .

$u^{2} - 2 u + 1^{2} = 0$

$x = \frac{1}{y}$

Step 3.3.3.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 u = 2 \cdot u \cdot 1$

$x = \frac{1}{y}$

Step 3.3.3.3

Rewrite the polynomial.

$u^{2} - 2 \cdot u \cdot 1 + 1^{2} = 0$

$x = \frac{1}{y}$

Step 3.3.3.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = u$ and $b = 1$ .

${(u - 1)}^{2} = 0$

$x = \frac{1}{y}$

${(u - 1)}^{2} = 0$

$x = \frac{1}{y}$

Step 3.3.4

Set the $u - 1$ equal to $0$ .

$u - 1 = 0$

$x = \frac{1}{y}$

Step 3.3.5

Add $1$ to both sides of the equation.

$u = 1$

$x = \frac{1}{y}$

Step 3.3.6

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

$y^{2} = 1$

$x = \frac{1}{y}$

Step 3.3.7

Solve the equation for $y$ .

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Step 3.3.7.1

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

$y = \pm \sqrt{1}$

$x = \frac{1}{y}$

Step 3.3.7.2

Any root of $1$ is $1$ .

$y = \pm 1$

$x = \frac{1}{y}$

Step 3.3.7.3

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

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Step 3.3.7.3.1

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

$y = 1$

$x = \frac{1}{y}$

Step 3.3.7.3.2

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

$y = - 1$

$x = \frac{1}{y}$

Step 3.3.7.3.3

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

$y = 1, - 1$

$x = \frac{1}{y}$

$y = 1, - 1$

$x = \frac{1}{y}$

$y = 1, - 1$

$x = \frac{1}{y}$

$y = 1, - 1$

$x = \frac{1}{y}$

$y = 1, - 1$

$x = \frac{1}{y}$

Step 4

Replace all occurrences of $y$ with $1$ in each equation.

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Step 4.1

Replace all occurrences of $y$ in $x = \frac{1}{y}$ with $1$ .

$x = \frac{1}{1}$

$y = 1$

Step 4.2

Simplify the right side.

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Step 4.2.1

Divide $1$ by $1$ .

$x = 1$

$y = 1$

$x = 1$

$y = 1$

$x = 1$

$y = 1$

Step 5

Replace all occurrences of $y$ with $- 1$ in each equation.

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Step 5.1

Replace all occurrences of $y$ in $x = \frac{1}{y}$ with $- 1$ .

$x = \frac{1}{- 1}$

$y = - 1$

Step 5.2

Simplify the right side.

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Step 5.2.1

Divide $1$ by $- 1$ .

$x = - 1$

$y = - 1$

$x = - 1$

$y = - 1$

$x = - 1$

$y = - 1$

Step 6

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

$(1, 1)$

$(- 1, - 1)$

Step 7

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 8