Algebra Examples

$\frac{5 z^{2}}{z^{2} + 5 z} \div \frac{B}{z^{2} - 2 z - 35} = 1$

Step 1

Factor each term.

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Step 1.1

Factor $z$ out of $z^{2} + 5 z$ .

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Step 1.1.1

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

$\frac{5 z^{2}}{z \cdot z + 5 z} \div \frac{B}{z^{2} - 2 z - 35} = 1$

Step 1.1.2

Factor $z$ out of $5 z$ .

$\frac{5 z^{2}}{z \cdot z + z \cdot 5} \div \frac{B}{z^{2} - 2 z - 35} = 1$

Step 1.1.3

Factor $z$ out of $z \cdot z + z \cdot 5$ .

$\frac{5 z^{2}}{z (z + 5)} \div \frac{B}{z^{2} - 2 z - 35} = 1$

Step 1.2

Reduce the expression $\frac{5 z^{2}}{z (z + 5)}$ by cancelling the common factors.

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Step 1.2.1

Factor $z$ out of $5 z^{2}$ .

$\frac{z (5 z)}{z (z + 5)} \div \frac{B}{z^{2} - 2 z - 35} = 1$

Step 1.2.2

Cancel the common factor.

$\frac{z (5 z)}{z (z + 5)} \div \frac{B}{z^{2} - 2 z - 35} = 1$

Step 1.2.3

Rewrite the expression.

$\frac{5 z}{z + 5} \div \frac{B}{z^{2} - 2 z - 35} = 1$

Step 1.3

Factor $z^{2} - 2 z - 35$ using the AC method.

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Step 1.3.1

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 $- 35$ and whose sum is $- 2$ .

$- 7, 5$

Step 1.3.2

Write the factored form using these integers.

$\frac{5 z}{z + 5} \div \frac{B}{(z - 7) (z + 5)} = 1$

Step 1.4

To divide by a fraction, multiply by its reciprocal.

$\frac{5 z}{z + 5} \cdot \frac{(z - 7) (z + 5)}{B} = 1$

Step 1.5

Cancel the common factor of $z + 5$ .

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Step 1.5.1

Factor $z + 5$ out of $(z - 7) (z + 5)$ .

$\frac{5 z}{z + 5} \cdot \frac{(z + 5) (z - 7)}{B} = 1$

Step 1.5.2

Cancel the common factor.

$\frac{5 z}{z + 5} \cdot \frac{(z + 5) (z - 7)}{B} = 1$

Step 1.5.3

Rewrite the expression.

$5 z \frac{z - 7}{B} = 1$

Step 1.6

Combine $5$ and $\frac{z - 7}{B}$ .

$z \frac{5 (z - 7)}{B} = 1$

Step 1.7

Combine $z$ and $\frac{5 (z - 7)}{B}$ .

$\frac{z (5 (z - 7))}{B} = 1$

Step 1.8

Remove unnecessary parentheses.

$\frac{z \cdot 5 (z - 7)}{B} = 1$

Step 1.9

Move $5$ to the left of $z$ .

$\frac{5 \cdot z (z - 7)}{B} = 1$

Step 1.10

Multiply $5$ by $z$ .

$\frac{5 z (z - 7)}{B} = 1$

Step 2

Find the LCD of the terms in the equation.

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Step 2.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$B, 1$

Step 2.2

The LCM of one and any expression is the expression.

$B$

Step 3

Multiply each term in $\frac{5 z (z - 7)}{B} = 1$ by $B$ to eliminate the fractions.

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Step 3.1

Multiply each term in $\frac{5 z (z - 7)}{B} = 1$ by $B$ .

$\frac{5 z (z - 7)}{B} B = 1 B$

Step 3.2

Simplify the left side.

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Step 3.2.1

Cancel the common factor of $B$ .

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Step 3.2.1.1

Cancel the common factor.

$\frac{5 z (z - 7)}{B} B = 1 B$

Step 3.2.1.2

Rewrite the expression.

$5 z (z - 7) = 1 B$

Step 3.2.2

Apply the distributive property.

$5 z \cdot z + 5 z \cdot - 7 = 1 B$

Step 3.2.3

Multiply $z$ by $z$ by adding the exponents.

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Step 3.2.3.1

Move $z$ .

$5 (z \cdot z) + 5 z \cdot - 7 = 1 B$

Step 3.2.3.2

Multiply $z$ by $z$ .

$5 z^{2} + 5 z \cdot - 7 = 1 B$

Step 3.2.4

Multiply $- 7$ by $5$ .

$5 z^{2} - 35 z = 1 B$

Step 3.3

Simplify the right side.

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Step 3.3.1

Multiply $B$ by $1$ .

$5 z^{2} - 35 z = B$

Step 4

Rewrite the equation as $B = 5 z^{2} - 35 z$ .

$B = 5 z^{2} - 35 z$