Precalculus Examples

$x y = 20$ , $2 x - y = - 6$

Step 1

Divide each term in $x y = 20$ by $y$ and simplify.

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Step 1.1

Divide each term in $x y = 20$ by $y$ .

$\frac{x y}{y} = \frac{20}{y}$

$2 x - y = - 6$

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.

$\frac{x y}{y} = \frac{20}{y}$

$2 x - y = - 6$

Step 1.2.1.2

Divide $x$ by $1$ .

$x = \frac{20}{y}$

$2 x - y = - 6$

$x = \frac{20}{y}$

$2 x - y = - 6$

$x = \frac{20}{y}$

$2 x - y = - 6$

$x = \frac{20}{y}$

$2 x - y = - 6$

Step 2

Replace all occurrences of $x$ with $\frac{20}{y}$ in each equation.

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Step 2.1

Replace all occurrences of $x$ in $2 x - y = - 6$ with $\frac{20}{y}$ .

$2 (\frac{20}{y}) - y = - 6$

$x = \frac{20}{y}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Multiply $2 (\frac{20}{y})$ .

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Step 2.2.1.1

Combine $2$ and $\frac{20}{y}$ .

$\frac{2 \cdot 20}{y} - y = - 6$

$x = \frac{20}{y}$

Step 2.2.1.2

Multiply $2$ by $20$ .

$\frac{40}{y} - y = - 6$

$x = \frac{20}{y}$

$\frac{40}{y} - y = - 6$

$x = \frac{20}{y}$

$\frac{40}{y} - y = - 6$

$x = \frac{20}{y}$

$\frac{40}{y} - y = - 6$

$x = \frac{20}{y}$

Step 3

Solve for $y$ in $\frac{40}{y} - y = - 6$ .

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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.

$y, 1, 1$

$x = \frac{20}{y}$

Step 3.1.2

The LCM of one and any expression is the expression.

$x = \frac{20}{y}$

Step 3.2

Multiply each term in $\frac{40}{y} - y = - 6$ by $y$ to eliminate the fractions.

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Step 3.2.1

Multiply each term in $\frac{40}{y} - y = - 6$ by $y$ .

$\frac{40}{y} \cdot y - y \cdot y = - 6 y$

$x = \frac{20}{y}$

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 $y$ .

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Step 3.2.2.1.1.1

Cancel the common factor.

$\frac{40}{y} \cdot y - y \cdot y = - 6 y$

$x = \frac{20}{y}$

Step 3.2.2.1.1.2

Rewrite the expression.

$40 - y \cdot y = - 6 y$

$x = \frac{20}{y}$

$40 - y \cdot y = - 6 y$

$x = \frac{20}{y}$

Step 3.2.2.1.2

Multiply $y$ by $y$ by adding the exponents.

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Step 3.2.2.1.2.1

Move $y$ .

$40 - (y \cdot y) = - 6 y$

$x = \frac{20}{y}$

Step 3.2.2.1.2.2

Multiply $y$ by $y$ .

$40 - y^{2} = - 6 y$

$x = \frac{20}{y}$

$40 - y^{2} = - 6 y$

$x = \frac{20}{y}$

$40 - y^{2} = - 6 y$

$x = \frac{20}{y}$

$40 - y^{2} = - 6 y$

$x = \frac{20}{y}$

$40 - y^{2} = - 6 y$

$x = \frac{20}{y}$

Step 3.3

Solve the equation.

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Step 3.3.1

Add $6 y$ to both sides of the equation.

$40 - y^{2} + 6 y = 0$

$x = \frac{20}{y}$

Step 3.3.2

Factor the left side of the equation.

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Step 3.3.2.1

Factor $- 1$ out of $40 - y^{2} + 6 y$ .

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Step 3.3.2.1.1

Move $40$ .

$- y^{2} + 6 y + 40 = 0$

$x = \frac{20}{y}$

Step 3.3.2.1.2

Factor $- 1$ out of $- y^{2}$ .

$- (y^{2}) + 6 y + 40 = 0$

$x = \frac{20}{y}$

Step 3.3.2.1.3

Factor $- 1$ out of $6 y$ .

$- (y^{2}) - (- 6 y) + 40 = 0$

$x = \frac{20}{y}$

Step 3.3.2.1.4

Rewrite $40$ as $- 1 (- 40)$ .

$- (y^{2}) - (- 6 y) - 1 \cdot - 40 = 0$

$x = \frac{20}{y}$

Step 3.3.2.1.5

Factor $- 1$ out of $- (y^{2}) - (- 6 y)$ .

$- (y^{2} - 6 y) - 1 \cdot - 40 = 0$

$x = \frac{20}{y}$

Step 3.3.2.1.6

Factor $- 1$ out of $- (y^{2} - 6 y) - 1 (- 40)$ .

$- (y^{2} - 6 y - 40) = 0$

$x = \frac{20}{y}$

$- (y^{2} - 6 y - 40) = 0$

$x = \frac{20}{y}$

Step 3.3.2.2

Factor.

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Step 3.3.2.2.1

Factor $y^{2} - 6 y - 40$ using the AC method.

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Step 3.3.2.2.1.1

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 $- 40$ and whose sum is $- 6$ .

$- 10, 4$

$x = \frac{20}{y}$

Step 3.3.2.2.1.2

Write the factored form using these integers.

$- ((y - 10) (y + 4)) = 0$

$x = \frac{20}{y}$

$- ((y - 10) (y + 4)) = 0$

$x = \frac{20}{y}$

Step 3.3.2.2.2

Remove unnecessary parentheses.

$- (y - 10) (y + 4) = 0$

$x = \frac{20}{y}$

$- (y - 10) (y + 4) = 0$

$x = \frac{20}{y}$

$- (y - 10) (y + 4) = 0$

$x = \frac{20}{y}$

Step 3.3.3

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

$y - 10 = 0$

$y + 4 = 0$

$x = \frac{20}{y}$

Step 3.3.4

Set $y - 10$ equal to $0$ and solve for $y$ .

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Step 3.3.4.1

Set $y - 10$ equal to $0$ .

$y - 10 = 0$

$x = \frac{20}{y}$

Step 3.3.4.2

Add $10$ to both sides of the equation.

$y = 10$

$x = \frac{20}{y}$

$y = 10$

$x = \frac{20}{y}$

Step 3.3.5

Set $y + 4$ equal to $0$ and solve for $y$ .

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Step 3.3.5.1

Set $y + 4$ equal to $0$ .

$y + 4 = 0$

$x = \frac{20}{y}$

Step 3.3.5.2

Subtract $4$ from both sides of the equation.

$y = - 4$

$x = \frac{20}{y}$

$y = - 4$

$x = \frac{20}{y}$

Step 3.3.6

The final solution is all the values that make $- (y - 10) (y + 4) = 0$ true.

$y = 10, - 4$

$x = \frac{20}{y}$

$y = 10, - 4$

$x = \frac{20}{y}$

$y = 10, - 4$

$x = \frac{20}{y}$

Step 4

Replace all occurrences of $y$ with $10$ in each equation.

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Step 4.1

Replace all occurrences of $y$ in $x = \frac{20}{y}$ with $10$ .

$x = \frac{20}{10}$

$y = 10$

Step 4.2

Simplify the right side.

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Step 4.2.1

Divide $20$ by $10$ .

$x = 2$

$y = 10$

$x = 2$

$y = 10$

$x = 2$

$y = 10$

Step 5

Replace all occurrences of $y$ with $- 4$ in each equation.

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Step 5.1

Replace all occurrences of $y$ in $x = \frac{20}{y}$ with $- 4$ .

$x = \frac{20}{- 4}$

$y = - 4$

Step 5.2

Simplify the right side.

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Step 5.2.1

Divide $20$ by $- 4$ .

$x = - 5$

$y = - 4$

$x = - 5$

$y = - 4$

$x = - 5$

$y = - 4$

Step 6

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

$(2, 10)$

$(- 5, - 4)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(2, 10), (- 5, - 4)$

Equation Form:

$x = 2, y = 10$

$x = - 5, y = - 4$

Step 8