Algebra Examples

$\frac{6}{t + 5} + \frac{2}{t - 5} = \frac{3 t - 1}{t^{2} - 25}$

Step 1

Subtract $\frac{3 t - 1}{t^{2} - 25}$ from both sides of the equation.

$\frac{6}{t + 5} + \frac{2}{t - 5} - \frac{3 t - 1}{t^{2} - 25} = 0$

Step 2

Simplify $\frac{6}{t + 5} + \frac{2}{t - 5} - \frac{3 t - 1}{t^{2} - 25}$ .

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Step 2.1

Simplify the denominator.

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Step 2.1.1

Rewrite $25$ as $5^{2}$ .

$\frac{6}{t + 5} + \frac{2}{t - 5} - \frac{3 t - 1}{t^{2} - 5^{2}} = 0$

Step 2.1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = t$ and $b = 5$ .

$\frac{6}{t + 5} + \frac{2}{t - 5} - \frac{3 t - 1}{(t + 5) (t - 5)} = 0$

Step 2.2

To write $\frac{6}{t + 5}$ as a fraction with a common denominator, multiply by $\frac{t - 5}{t - 5}$ .

$\frac{6}{t + 5} \cdot \frac{t - 5}{t - 5} + \frac{2}{t - 5} - \frac{3 t - 1}{(t + 5) (t - 5)} = 0$

Step 2.3

To write $\frac{2}{t - 5}$ as a fraction with a common denominator, multiply by $\frac{t + 5}{t + 5}$ .

$\frac{6}{t + 5} \cdot \frac{t - 5}{t - 5} + \frac{2}{t - 5} \cdot \frac{t + 5}{t + 5} - \frac{3 t - 1}{(t + 5) (t - 5)} = 0$

Step 2.4

Write each expression with a common denominator of $(t + 5) (t - 5)$ , by multiplying each by an appropriate factor of $1$ .

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Step 2.4.1

Multiply $\frac{6}{t + 5}$ by $\frac{t - 5}{t - 5}$ .

$\frac{6 (t - 5)}{(t + 5) (t - 5)} + \frac{2}{t - 5} \cdot \frac{t + 5}{t + 5} - \frac{3 t - 1}{(t + 5) (t - 5)} = 0$

Step 2.4.2

Multiply $\frac{2}{t - 5}$ by $\frac{t + 5}{t + 5}$ .

$\frac{6 (t - 5)}{(t + 5) (t - 5)} + \frac{2 (t + 5)}{(t - 5) (t + 5)} - \frac{3 t - 1}{(t + 5) (t - 5)} = 0$

Step 2.4.3

Reorder the factors of $(t - 5) (t + 5)$ .

$\frac{6 (t - 5)}{(t + 5) (t - 5)} + \frac{2 (t + 5)}{(t + 5) (t - 5)} - \frac{3 t - 1}{(t + 5) (t - 5)} = 0$

Step 2.5

Combine the numerators over the common denominator.

$\frac{6 (t - 5) + 2 (t + 5)}{(t + 5) (t - 5)} - \frac{3 t - 1}{(t + 5) (t - 5)} = 0$

Step 2.6

Combine the numerators over the common denominator.

$\frac{6 (t - 5) + 2 (t + 5) - (3 t - 1)}{(t + 5) (t - 5)} = 0$

Step 2.7

Simplify each term.

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Step 2.7.1

Apply the distributive property.

$\frac{6 t + 6 \cdot - 5 + 2 (t + 5) - (3 t - 1)}{(t + 5) (t - 5)} = 0$

Step 2.7.2

Multiply $6$ by $- 5$ .

$\frac{6 t - 30 + 2 (t + 5) - (3 t - 1)}{(t + 5) (t - 5)} = 0$

Step 2.7.3

Apply the distributive property.

$\frac{6 t - 30 + 2 t + 2 \cdot 5 - (3 t - 1)}{(t + 5) (t - 5)} = 0$

Step 2.7.4

Multiply $2$ by $5$ .

$\frac{6 t - 30 + 2 t + 10 - (3 t - 1)}{(t + 5) (t - 5)} = 0$

Step 2.7.5

Apply the distributive property.

$\frac{6 t - 30 + 2 t + 10 - (3 t) - - 1}{(t + 5) (t - 5)} = 0$

Step 2.7.6

Multiply $3$ by $- 1$ .

$\frac{6 t - 30 + 2 t + 10 - 3 t - - 1}{(t + 5) (t - 5)} = 0$

Step 2.7.7

Multiply $- 1$ by $- 1$ .

$\frac{6 t - 30 + 2 t + 10 - 3 t + 1}{(t + 5) (t - 5)} = 0$

Step 2.8

Add $6 t$ and $2 t$ .

$\frac{8 t - 30 + 10 - 3 t + 1}{(t + 5) (t - 5)} = 0$

Step 2.9

Subtract $3 t$ from $8 t$ .

$\frac{5 t - 30 + 10 + 1}{(t + 5) (t - 5)} = 0$

Step 2.10

Add $- 30$ and $10$ .

$\frac{5 t - 20 + 1}{(t + 5) (t - 5)} = 0$

Step 2.11

Add $- 20$ and $1$ .

$\frac{5 t - 19}{(t + 5) (t - 5)} = 0$

Step 3

Set the denominator in $\frac{5 t - 19}{(t + 5) (t - 5)}$ equal to $0$ to find where the expression is undefined.

$(t + 5) (t - 5) = 0$

Step 4

Solve for $t$ .

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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 $0$ .

$t + 5 = 0$

$t - 5 = 0$

Step 4.2

Set $t + 5$ equal to $0$ and solve for $t$ .

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Step 4.2.1

Set $t + 5$ equal to $0$ .

$t + 5 = 0$

Step 4.2.2

Subtract $5$ from both sides of the equation.

$t = - 5$

Step 4.3

Set $t - 5$ equal to $0$ and solve for $t$ .

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Step 4.3.1

Set $t - 5$ equal to $0$ .

$t - 5 = 0$

Step 4.3.2

Add $5$ to both sides of the equation.

$t = 5$

Step 4.4

The final solution is all the values that make $(t + 5) (t - 5) = 0$ true.

$t = - 5, 5$

Step 5

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$t = - 5, t = 5$

Step 6