Finite Math Examples

$13 - \frac{20}{y} = y + 22$

Step 1

Move all terms not containing $y$ to the right side of the equation.

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Step 1.1

Subtract $13$ from both sides of the equation.

$- \frac{20}{y} = y + 22 - 13$

Step 1.2

Subtract $13$ from $22$ .

$- \frac{20}{y} = y + 9$

Step 2

Find the LCD of the terms in the equation.

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Step 2.1

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

$y, 1, 1$

Step 2.2

The LCM of one and any expression is the expression.

$y$

Step 3

Multiply each term in $- \frac{20}{y} = y + 9$ by $y$ to eliminate the fractions.

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Step 3.1

Multiply each term in $- \frac{20}{y} = y + 9$ by $y$ .

$- \frac{20}{y} y = y \cdot y + 9 y$

Step 3.2

Simplify the left side.

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Step 3.2.1

Cancel the common factor of $y$ .

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Step 3.2.1.1

Move the leading negative in $- \frac{20}{y}$ into the numerator.

$\frac{- 20}{y} y = y \cdot y + 9 y$

Step 3.2.1.2

Cancel the common factor.

$\frac{- 20}{y} y = y \cdot y + 9 y$

Step 3.2.1.3

Rewrite the expression.

$- 20 = y \cdot y + 9 y$

Step 3.3

Simplify the right side.

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Step 3.3.1

Multiply $y$ by $y$ .

$- 20 = y^{2} + 9 y$

Step 4

Solve the equation.

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Step 4.1

Rewrite the equation as $y^{2} + 9 y = - 20$ .

$y^{2} + 9 y = - 20$

Step 4.2

Add $20$ to both sides of the equation.

$y^{2} + 9 y + 20 = 0$

Step 4.3

Factor $y^{2} + 9 y + 20$ using the AC method.

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Step 4.3.1

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 $20$ and whose sum is $9$ .

$4, 5$

Step 4.3.2

Write the factored form using these integers.

$(y + 4) (y + 5) = 0$

Step 4.4

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

$y + 4 = 0$

$y + 5 = 0$

Step 4.5

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

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Step 4.5.1

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

$y + 4 = 0$

Step 4.5.2

Subtract $4$ from both sides of the equation.

$y = - 4$

Step 4.6

Set $y + 5$ equal to $0$ and solve for $y$ .

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Step 4.6.1

Set $y + 5$ equal to $0$ .

$y + 5 = 0$

Step 4.6.2

Subtract $5$ from both sides of the equation.

$y = - 5$

Step 4.7

The final solution is all the values that make $(y + 4) (y + 5) = 0$ true.

$y = - 4, - 5$