Basic Math Examples

$\frac{1}{{(q + 3)}^{2}} = \frac{1}{q + 3} + 12$

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.

${(q + 3)}^{2}, q + 3, 1$

Step 1.2

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Step 1.3

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

Step 1.4

The LCM of $1, 1, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$1$

Step 1.5

The factors for $q + 3$ are $(q + 3) \cdot (q + 3)$ , which is $q + 3$ multiplied by itself $2$ times.

$(q + 3) = (q + 3) \cdot (q + 3)$

$(q + 3)$ occurs $2$ times.

Step 1.6

The factor for $q + 3$ is $q + 3$ itself.

$(q + 3) = q + 3$

$(q + 3)$ occurs $1$ time.

Step 1.7

The LCM of ${(q + 3)}^{2}, q + 3$ is the result of multiplying all factors the greatest number of times they occur in either term.

${(q + 3)}^{2}$

Step 2

Multiply each term in $\frac{1}{{(q + 3)}^{2}} = \frac{1}{q + 3} + 12$ by ${(q + 3)}^{2}$ to eliminate the fractions.

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Step 2.1

Multiply each term in $\frac{1}{{(q + 3)}^{2}} = \frac{1}{q + 3} + 12$ by ${(q + 3)}^{2}$ .

$\frac{1}{{(q + 3)}^{2}} {(q + 3)}^{2} = \frac{1}{q + 3} {(q + 3)}^{2} + 12 {(q + 3)}^{2}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Cancel the common factor of ${(q + 3)}^{2}$ .

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Step 2.2.1.1

Cancel the common factor.

$\frac{1}{{(q + 3)}^{2}} {(q + 3)}^{2} = \frac{1}{q + 3} {(q + 3)}^{2} + 12 {(q + 3)}^{2}$

Step 2.2.1.2

Rewrite the expression.

$1 = \frac{1}{q + 3} {(q + 3)}^{2} + 12 {(q + 3)}^{2}$

Step 2.3

Simplify the right side.

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Step 2.3.1

Cancel the common factor of $q + 3$ .

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Step 2.3.1.1

Factor $q + 3$ out of ${(q + 3)}^{2}$ .

$1 = \frac{1}{q + 3} ((q + 3) (q + 3)) + 12 {(q + 3)}^{2}$

Step 2.3.1.2

Cancel the common factor.

$1 = \frac{1}{q + 3} ((q + 3) (q + 3)) + 12 {(q + 3)}^{2}$

Step 2.3.1.3

Rewrite the expression.

$1 = q + 3 + 12 {(q + 3)}^{2}$

Step 3

Solve the equation.

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Step 3.1

Rewrite the equation as $q + 3 + 12 {(q + 3)}^{2} = 1$ .

$q + 3 + 12 {(q + 3)}^{2} = 1$

Step 3.2

Subtract $1$ from both sides of the equation.

$q + 3 + 12 {(q + 3)}^{2} - 1 = 0$

Step 3.3

Simplify $q + 3 + 12 {(q + 3)}^{2} - 1$ .

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Step 3.3.1

Simplify each term.

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Step 3.3.1.1

Rewrite ${(q + 3)}^{2}$ as $(q + 3) (q + 3)$ .

$q + 3 + 12 ((q + 3) (q + 3)) - 1 = 0$

Step 3.3.1.2

Expand $(q + 3) (q + 3)$ using the FOIL Method.

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Step 3.3.1.2.1

Apply the distributive property.

$q + 3 + 12 (q (q + 3) + 3 (q + 3)) - 1 = 0$

Step 3.3.1.2.2

Apply the distributive property.

$q + 3 + 12 (q \cdot q + q \cdot 3 + 3 (q + 3)) - 1 = 0$

Step 3.3.1.2.3

Apply the distributive property.

$q + 3 + 12 (q \cdot q + q \cdot 3 + 3 q + 3 \cdot 3) - 1 = 0$

Step 3.3.1.3

Simplify and combine like terms.

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Step 3.3.1.3.1

Simplify each term.

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Step 3.3.1.3.1.1

Multiply $q$ by $q$ .

$q + 3 + 12 (q^{2} + q \cdot 3 + 3 q + 3 \cdot 3) - 1 = 0$

Step 3.3.1.3.1.2

Move $3$ to the left of $q$ .

$q + 3 + 12 (q^{2} + 3 \cdot q + 3 q + 3 \cdot 3) - 1 = 0$

Step 3.3.1.3.1.3

Multiply $3$ by $3$ .

$q + 3 + 12 (q^{2} + 3 q + 3 q + 9) - 1 = 0$

Step 3.3.1.3.2

Add $3 q$ and $3 q$ .

$q + 3 + 12 (q^{2} + 6 q + 9) - 1 = 0$

Step 3.3.1.4

Apply the distributive property.

$q + 3 + 12 q^{2} + 12 (6 q) + 12 \cdot 9 - 1 = 0$

Step 3.3.1.5

Simplify.

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Step 3.3.1.5.1

Multiply $6$ by $12$ .

$q + 3 + 12 q^{2} + 72 q + 12 \cdot 9 - 1 = 0$

Step 3.3.1.5.2

Multiply $12$ by $9$ .

$q + 3 + 12 q^{2} + 72 q + 108 - 1 = 0$

Step 3.3.2

Add $q$ and $72 q$ .

$73 q + 3 + 12 q^{2} + 108 - 1 = 0$

Step 3.3.3

Add $3$ and $108$ .

$73 q + 12 q^{2} + 111 - 1 = 0$

Step 3.3.4

Subtract $1$ from $111$ .

$73 q + 12 q^{2} + 110 = 0$

Step 3.4

Factor the left side of the equation.

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Step 3.4.1

Let $u = q$ . Substitute $u$ for all occurrences of $q$ .

$73 u + 12 u^{2} + 110 = 0$

Step 3.4.2

Factor by grouping.

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Step 3.4.2.1

Reorder terms.

$12 u^{2} + 73 u + 110 = 0$

Step 3.4.2.2

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 12 \cdot 110 = 1320$ and whose sum is $b = 73$ .

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Step 3.4.2.2.1

Factor $73$ out of $73 u$ .

$12 u^{2} + 73 (u) + 110 = 0$

Step 3.4.2.2.2

Rewrite $73$ as $33$ plus $40$

$12 u^{2} + (33 + 40) u + 110 = 0$

Step 3.4.2.2.3

Apply the distributive property.

$12 u^{2} + 33 u + 40 u + 110 = 0$

Step 3.4.2.3

Factor out the greatest common factor from each group.

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Step 3.4.2.3.1

Group the first two terms and the last two terms.

$(12 u^{2} + 33 u) + 40 u + 110 = 0$

Step 3.4.2.3.2

Factor out the greatest common factor (GCF) from each group.

$3 u (4 u + 11) + 10 (4 u + 11) = 0$

Step 3.4.2.4

Factor the polynomial by factoring out the greatest common factor, $4 u + 11$ .

$(4 u + 11) (3 u + 10) = 0$

Step 3.4.3

Replace all occurrences of $u$ with $q$ .

$(4 q + 11) (3 q + 10) = 0$

Step 3.5

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

$4 q + 11 = 0$

$3 q + 10 = 0$

Step 3.6

Set $4 q + 11$ equal to $0$ and solve for $q$ .

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Step 3.6.1

Set $4 q + 11$ equal to $0$ .

$4 q + 11 = 0$

Step 3.6.2

Solve $4 q + 11 = 0$ for $q$ .

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Step 3.6.2.1

Subtract $11$ from both sides of the equation.

$4 q = - 11$

Step 3.6.2.2

Divide each term in $4 q = - 11$ by $4$ and simplify.

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Step 3.6.2.2.1

Divide each term in $4 q = - 11$ by $4$ .

$\frac{4 q}{4} = \frac{- 11}{4}$

Step 3.6.2.2.2

Simplify the left side.

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Step 3.6.2.2.2.1

Cancel the common factor of $4$ .

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Step 3.6.2.2.2.1.1

Cancel the common factor.

$\frac{4 q}{4} = \frac{- 11}{4}$

Step 3.6.2.2.2.1.2

Divide $q$ by $1$ .

$q = \frac{- 11}{4}$

Step 3.6.2.2.3

Simplify the right side.

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Step 3.6.2.2.3.1

Move the negative in front of the fraction.

$q = - \frac{11}{4}$

Step 3.7

Set $3 q + 10$ equal to $0$ and solve for $q$ .

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Step 3.7.1

Set $3 q + 10$ equal to $0$ .

$3 q + 10 = 0$

Step 3.7.2

Solve $3 q + 10 = 0$ for $q$ .

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Step 3.7.2.1

Subtract $10$ from both sides of the equation.

$3 q = - 10$

Step 3.7.2.2

Divide each term in $3 q = - 10$ by $3$ and simplify.

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Step 3.7.2.2.1

Divide each term in $3 q = - 10$ by $3$ .

$\frac{3 q}{3} = \frac{- 10}{3}$

Step 3.7.2.2.2

Simplify the left side.

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Step 3.7.2.2.2.1

Cancel the common factor of $3$ .

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Step 3.7.2.2.2.1.1

Cancel the common factor.

$\frac{3 q}{3} = \frac{- 10}{3}$

Step 3.7.2.2.2.1.2

Divide $q$ by $1$ .

$q = \frac{- 10}{3}$

Step 3.7.2.2.3

Simplify the right side.

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Step 3.7.2.2.3.1

Move the negative in front of the fraction.

$q = - \frac{10}{3}$

Step 3.8

The final solution is all the values that make $(4 q + 11) (3 q + 10) = 0$ true.

$q = - \frac{11}{4}, - \frac{10}{3}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$q = - \frac{11}{4}, - \frac{10}{3}$

Decimal Form:

$q = - 2.75, - 3.\overline{3}$

Mixed Number Form:

$q = - 2 \frac{3}{4}, - 3 \frac{1}{3}$