Algebra Examples

$\frac{p + 5}{p^{2} + p} = \frac{1}{p^{2} + p} - \frac{p - 6}{p + 1}$

Step 1

Move all the expressions to the left side of the equation.

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Step 1.1

Subtract $\frac{1}{p^{2} + p}$ from both sides of the equation.

$\frac{p + 5}{p^{2} + p} - \frac{1}{p^{2} + p} = - \frac{p - 6}{p + 1}$

Step 1.2

Add $\frac{p - 6}{p + 1}$ to both sides of the equation.

$\frac{p + 5}{p^{2} + p} - \frac{1}{p^{2} + p} + \frac{p - 6}{p + 1} = 0$

Step 2

Simplify $\frac{p + 5}{p^{2} + p} - \frac{1}{p^{2} + p} + \frac{p - 6}{p + 1}$ .

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Step 2.1

Combine the numerators over the common denominator.

$\frac{p + 5 - 1}{p^{2} + p} + \frac{p - 6}{p + 1} = 0$

Step 2.2

Subtract $1$ from $5$ .

$\frac{p + 4}{p^{2} + p} + \frac{p - 6}{p + 1} = 0$

Step 2.3

Factor $p$ out of $p^{2} + p$ .

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Step 2.3.1

Factor $p$ out of $p^{2}$ .

$\frac{p + 4}{p \cdot p + p} + \frac{p - 6}{p + 1} = 0$

Step 2.3.2

Raise $p$ to the power of $1$ .

$\frac{p + 4}{p \cdot p + p^{1}} + \frac{p - 6}{p + 1} = 0$

Step 2.3.3

Factor $p$ out of $p^{1}$ .

$\frac{p + 4}{p \cdot p + p \cdot 1} + \frac{p - 6}{p + 1} = 0$

Step 2.3.4

Factor $p$ out of $p \cdot p + p \cdot 1$ .

$\frac{p + 4}{p (p + 1)} + \frac{p - 6}{p + 1} = 0$

Step 2.4

To write $\frac{p - 6}{p + 1}$ as a fraction with a common denominator, multiply by $\frac{p}{p}$ .

$\frac{p + 4}{p (p + 1)} + \frac{p - 6}{p + 1} \cdot \frac{p}{p} = 0$

Step 2.5

Write each expression with a common denominator of $p (p + 1)$ , by multiplying each by an appropriate factor of $1$ .

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Step 2.5.1

Multiply $\frac{p - 6}{p + 1}$ by $\frac{p}{p}$ .

$\frac{p + 4}{p (p + 1)} + \frac{(p - 6) p}{(p + 1) p} = 0$

Step 2.5.2

Reorder the factors of $(p + 1) p$ .

$\frac{p + 4}{p (p + 1)} + \frac{(p - 6) p}{p (p + 1)} = 0$

Step 2.6

Combine the numerators over the common denominator.

$\frac{p + 4 + (p - 6) p}{p (p + 1)} = 0$

Step 2.7

Simplify the numerator.

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Step 2.7.1

Apply the distributive property.

$\frac{p + 4 + p \cdot p - 6 p}{p (p + 1)} = 0$

Step 2.7.2

Multiply $p$ by $p$ .

$\frac{p + 4 + p^{2} - 6 p}{p (p + 1)} = 0$

Step 2.7.3

Subtract $6 p$ from $p$ .

$\frac{- 5 p + 4 + p^{2}}{p (p + 1)} = 0$

Step 2.7.4

Reorder terms.

$\frac{p^{2} - 5 p + 4}{p (p + 1)} = 0$

Step 2.7.5

Factor $p^{2} - 5 p + 4$ using the AC method.

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Step 2.7.5.1

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 $4$ and whose sum is $- 5$ .

$- 4, - 1$

Step 2.7.5.2

Write the factored form using these integers.

$\frac{(p - 4) (p - 1)}{p (p + 1)} = 0$

Step 3

Set the numerator equal to zero.

$(p - 4) (p - 1) = 0$

Step 4

Solve the equation for $p$ .

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Step 4.1

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

$p - 4 = 0$

$p - 1 = 0$

Step 4.2

Set $p - 4$ equal to $0$ and solve for $p$ .

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Step 4.2.1

Set $p - 4$ equal to $0$ .

$p - 4 = 0$

Step 4.2.2

Add $4$ to both sides of the equation.

$p = 4$

Step 4.3

Set $p - 1$ equal to $0$ and solve for $p$ .

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Step 4.3.1

Set $p - 1$ equal to $0$ .

$p - 1 = 0$

Step 4.3.2

Add $1$ to both sides of the equation.

$p = 1$

Step 4.4

The final solution is all the values that make $(p - 4) (p - 1) = 0$ true.

$p = 4, 1$