Precalculus Examples

$y + \frac{6}{y} = 5$

Step 1

Find the LCD of the terms in the equation.

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Step 1.1

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

$1, y, 1$

Step 1.2

The LCM of one and any expression is the expression.

$y$

Step 2

Multiply each term in $y + \frac{6}{y} = 5$ by $y$ to eliminate the fractions.

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Step 2.1

Multiply each term in $y + \frac{6}{y} = 5$ by $y$ .

$y \cdot y + \frac{6}{y} y = 5 y$

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

Multiply $y$ by $y$ .

$y^{2} + \frac{6}{y} y = 5 y$

Step 2.2.1.2

Cancel the common factor of $y$ .

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Step 2.2.1.2.1

Cancel the common factor.

$y^{2} + \frac{6}{y} y = 5 y$

Step 2.2.1.2.2

Rewrite the expression.

$y^{2} + 6 = 5 y$

Step 3

Solve the equation.

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Step 3.1

Subtract $5 y$ from both sides of the equation.

$y^{2} + 6 - 5 y = 0$

Step 3.2

Factor $y^{2} + 6 - 5 y$ using the AC method.

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Step 3.2.1

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 $6$ and whose sum is $- 5$ .

$- 3, - 2$

Step 3.2.2

Write the factored form using these integers.

$(y - 3) (y - 2) = 0$

Step 3.3

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

$y - 3 = 0$

$y - 2 = 0$

Step 3.4

Set $y - 3$ equal to $0$ and solve for $y$ .

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Step 3.4.1

Set $y - 3$ equal to $0$ .

$y - 3 = 0$

Step 3.4.2

Add $3$ to both sides of the equation.

$y = 3$

Step 3.5

Set $y - 2$ equal to $0$ and solve for $y$ .

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Step 3.5.1

Set $y - 2$ equal to $0$ .

$y - 2 = 0$

Step 3.5.2

Add $2$ to both sides of the equation.

$y = 2$

Step 3.6

The final solution is all the values that make $(y - 3) (y - 2) = 0$ true.

$y = 3, 2$