Pre-Algebra Examples

$q + \frac{6}{q} = - 7$

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.

$1, q, 1$

Step 1.2

The LCM of one and any expression is the expression.

$q$

Step 2

Multiply each term in $q + \frac{6}{q} = - 7$ by $q$ to eliminate the fractions.

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Step 2.1

Multiply each term in $q + \frac{6}{q} = - 7$ by $q$ .

$q \cdot q + \frac{6}{q} q = - 7 q$

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 $q$ by $q$ .

$q^{2} + \frac{6}{q} q = - 7 q$

Step 2.2.1.2

Cancel the common factor of $q$ .

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Step 2.2.1.2.1

Cancel the common factor.

$q^{2} + \frac{6}{q} q = - 7 q$

Step 2.2.1.2.2

Rewrite the expression.

$q^{2} + 6 = - 7 q$

Step 3

Solve the equation.

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Step 3.1

Add $7 q$ to both sides of the equation.

$q^{2} + 6 + 7 q = 0$

Step 3.2

Factor $q^{2} + 6 + 7 q$ 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 $7$ .

$1, 6$

Step 3.2.2

Write the factored form using these integers.

$(q + 1) (q + 6) = 0$

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 $0$ .

$q + 1 = 0$

$q + 6 = 0$

Step 3.4

Set $q + 1$ equal to $0$ and solve for $q$ .

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Step 3.4.1

Set $q + 1$ equal to $0$ .

$q + 1 = 0$

Step 3.4.2

Subtract $1$ from both sides of the equation.

$q = - 1$

Step 3.5

Set $q + 6$ equal to $0$ and solve for $q$ .

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Step 3.5.1

Set $q + 6$ equal to $0$ .

$q + 6 = 0$

Step 3.5.2

Subtract $6$ from both sides of the equation.

$q = - 6$

Step 3.6

The final solution is all the values that make $(q + 1) (q + 6) = 0$ true.

$q = - 1, - 6$