Precalculus Examples

$\frac{12}{t} + t - 8 = 0$

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.

$t, 1, 1, 1$

Step 1.2

The LCM of one and any expression is the expression.

$t$

Step 2

Multiply each term in $\frac{12}{t} + t - 8 = 0$ by $t$ to eliminate the fractions.

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Step 2.1

Multiply each term in $\frac{12}{t} + t - 8 = 0$ by $t$ .

$\frac{12}{t} t + t \cdot t - 8 t = 0 t$

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

Cancel the common factor of $t$ .

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Step 2.2.1.1.1

Cancel the common factor.

$\frac{12}{t} t + t \cdot t - 8 t = 0 t$

Step 2.2.1.1.2

Rewrite the expression.

$12 + t \cdot t - 8 t = 0 t$

Step 2.2.1.2

Multiply $t$ by $t$ .

$12 + t^{2} - 8 t = 0 t$

Step 2.3

Simplify the right side.

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Step 2.3.1

Multiply $0$ by $t$ .

$12 + t^{2} - 8 t = 0$

Step 3

Solve the equation.

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Step 3.1

Factor $12 + t^{2} - 8 t$ using the AC method.

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Step 3.1.1

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 $12$ and whose sum is $- 8$ .

$- 6, - 2$

Step 3.1.2

Write the factored form using these integers.

$(t - 6) (t - 2) = 0$

Step 3.2

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

$t - 6 = 0$

$t - 2 = 0$

Step 3.3

Set $t - 6$ equal to $0$ and solve for $t$ .

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Step 3.3.1

Set $t - 6$ equal to $0$ .

$t - 6 = 0$

Step 3.3.2

Add $6$ to both sides of the equation.

$t = 6$

Step 3.4

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

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Step 3.4.1

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

$t - 2 = 0$

Step 3.4.2

Add $2$ to both sides of the equation.

$t = 2$

Step 3.5

The final solution is all the values that make $(t - 6) (t - 2) = 0$ true.

$t = 6, 2$