Basic Math Examples

$\frac{1}{2} + \frac{2}{z + 1} = \frac{3}{4}$

Step 1

Move all terms not containing $z$ to the right side of the equation.

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Step 1.1

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

$\frac{2}{z + 1} = \frac{3}{4} - \frac{1}{2}$

Step 1.2

To write $- \frac{1}{2}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{2}{z + 1} = \frac{3}{4} - \frac{1}{2} \cdot \frac{2}{2}$

Step 1.3

Write each expression with a common denominator of $4$ , by multiplying each by an appropriate factor of $1$ .

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Step 1.3.1

Multiply $\frac{1}{2}$ by $\frac{2}{2}$ .

$\frac{2}{z + 1} = \frac{3}{4} - \frac{2}{2 \cdot 2}$

Step 1.3.2

Multiply $2$ by $2$ .

$\frac{2}{z + 1} = \frac{3}{4} - \frac{2}{4}$

Step 1.4

Combine the numerators over the common denominator.

$\frac{2}{z + 1} = \frac{3 - 2}{4}$

Step 1.5

Subtract $2$ from $3$ .

$\frac{2}{z + 1} = \frac{1}{4}$

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.

$z + 1, 4$

Step 2.2

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 2.3

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

Not prime

Step 2.4

$4$ has factors of $2$ and $2$ .

$2 \cdot 2$

Step 2.5

Multiply $2$ by $2$ .

$4$

Step 2.6

The factor for $z + 1$ is $z + 1$ itself.

$(z + 1) = z + 1$

$(z + 1)$ occurs $1$ time.

Step 2.7

The LCM of $z + 1$ is the result of multiplying all factors the greatest number of times they occur in either term.

$z + 1$

Step 2.8

The Least Common Multiple $LCM$ of some numbers is the smallest number that the numbers are factors of.

$4 (z + 1)$

Step 3

Multiply each term in $\frac{2}{z + 1} = \frac{1}{4}$ by $4 (z + 1)$ to eliminate the fractions.

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Step 3.1

Multiply each term in $\frac{2}{z + 1} = \frac{1}{4}$ by $4 (z + 1)$ .

$\frac{2}{z + 1} (4 (z + 1)) = \frac{1}{4} (4 (z + 1))$

Step 3.2

Simplify the left side.

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Step 3.2.1

Rewrite using the commutative property of multiplication.

$4 \frac{2}{z + 1} (z + 1) = \frac{1}{4} (4 (z + 1))$

Step 3.2.2

Multiply $4 \frac{2}{z + 1}$ .

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Step 3.2.2.1

Combine $4$ and $\frac{2}{z + 1}$ .

$\frac{4 \cdot 2}{z + 1} (z + 1) = \frac{1}{4} (4 (z + 1))$

Step 3.2.2.2

Multiply $4$ by $2$ .

$\frac{8}{z + 1} (z + 1) = \frac{1}{4} (4 (z + 1))$

Step 3.2.3

Cancel the common factor of $z + 1$ .

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Step 3.2.3.1

Cancel the common factor.

$\frac{8}{z + 1} (z + 1) = \frac{1}{4} (4 (z + 1))$

Step 3.2.3.2

Rewrite the expression.

$8 = \frac{1}{4} (4 (z + 1))$

Step 3.3

Simplify the right side.

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Step 3.3.1

Cancel the common factor of $4$ .

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Step 3.3.1.1

Cancel the common factor.

$8 = \frac{1}{4} (4 (z + 1))$

Step 3.3.1.2

Rewrite the expression.

$8 = z + 1$

Step 4

Solve the equation.

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Step 4.1

Rewrite the equation as $z + 1 = 8$ .

$z + 1 = 8$

Step 4.2

Move all terms not containing $z$ to the right side of the equation.

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Step 4.2.1

Subtract $1$ from both sides of the equation.

$z = 8 - 1$

Step 4.2.2

Subtract $1$ from $8$ .

$z = 7$