Precalculus Examples

$\frac{2 y + 1}{5} - \frac{2 + 7 y}{15} > \frac{2}{3}$

Step 1

Simplify $\frac{2 y + 1}{5} - \frac{2 + 7 y}{15}$ .

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Step 1.1

To write $\frac{2 y + 1}{5}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{2 y + 1}{5} \cdot \frac{3}{3} - \frac{2 + 7 y}{15} > \frac{2}{3}$

Step 1.2

Write each expression with a common denominator of $15$ , by multiplying each by an appropriate factor of $1$ .

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Step 1.2.1

Multiply $\frac{2 y + 1}{5}$ by $\frac{3}{3}$ .

$\frac{(2 y + 1) \cdot 3}{5 \cdot 3} - \frac{2 + 7 y}{15} > \frac{2}{3}$

Step 1.2.2

Multiply $5$ by $3$ .

$\frac{(2 y + 1) \cdot 3}{15} - \frac{2 + 7 y}{15} > \frac{2}{3}$

Step 1.3

Combine the numerators over the common denominator.

$\frac{(2 y + 1) \cdot 3 - (2 + 7 y)}{15} > \frac{2}{3}$

Step 1.4

Simplify the numerator.

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Step 1.4.1

Apply the distributive property.

$\frac{2 y \cdot 3 + 1 \cdot 3 - (2 + 7 y)}{15} > \frac{2}{3}$

Step 1.4.2

Multiply $3$ by $2$ .

$\frac{6 y + 1 \cdot 3 - (2 + 7 y)}{15} > \frac{2}{3}$

Step 1.4.3

Multiply $3$ by $1$ .

$\frac{6 y + 3 - (2 + 7 y)}{15} > \frac{2}{3}$

Step 1.4.4

Apply the distributive property.

$\frac{6 y + 3 - 1 \cdot 2 - (7 y)}{15} > \frac{2}{3}$

Step 1.4.5

Multiply $- 1$ by $2$ .

$\frac{6 y + 3 - 2 - (7 y)}{15} > \frac{2}{3}$

Step 1.4.6

Multiply $7$ by $- 1$ .

$\frac{6 y + 3 - 2 - 7 y}{15} > \frac{2}{3}$

Step 1.4.7

Subtract $7 y$ from $6 y$ .

$\frac{- y + 3 - 2}{15} > \frac{2}{3}$

Step 1.4.8

Subtract $2$ from $3$ .

$\frac{- y + 1}{15} > \frac{2}{3}$

Step 1.5

Simplify with factoring out.

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Step 1.5.1

Factor $- 1$ out of $- y$ .

$\frac{- (y) + 1}{15} > \frac{2}{3}$

Step 1.5.2

Rewrite $1$ as $- 1 (- 1)$ .

$\frac{- (y) - 1 (- 1)}{15} > \frac{2}{3}$

Step 1.5.3

Factor $- 1$ out of $- (y) - 1 (- 1)$ .

$\frac{- (y - 1)}{15} > \frac{2}{3}$

Step 1.5.4

Simplify the expression.

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Step 1.5.4.1

Rewrite $- (y - 1)$ as $- 1 (y - 1)$ .

$\frac{- 1 (y - 1)}{15} > \frac{2}{3}$

Step 1.5.4.2

Move the negative in front of the fraction.

$- \frac{y - 1}{15} > \frac{2}{3}$

Step 2

Divide each term in $- \frac{y - 1}{15} > \frac{2}{3}$ by $- 1$ and simplify.

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Step 2.1

Divide each term in $- \frac{y - 1}{15} > \frac{2}{3}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- \frac{y - 1}{15}}{- 1} < \frac{\frac{2}{3}}{- 1}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Dividing two negative values results in a positive value.

$\frac{\frac{y - 1}{15}}{1} < \frac{\frac{2}{3}}{- 1}$

Step 2.2.2

Divide $\frac{y - 1}{15}$ by $1$ .

$\frac{y - 1}{15} < \frac{\frac{2}{3}}{- 1}$

Step 2.3

Simplify the right side.

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Step 2.3.1

Move the negative one from the denominator of $\frac{\frac{2}{3}}{- 1}$ .

$\frac{y - 1}{15} < - 1 \cdot \frac{2}{3}$

Step 2.3.2

Rewrite $- 1 \cdot \frac{2}{3}$ as $- \frac{2}{3}$ .

$\frac{y - 1}{15} < - \frac{2}{3}$

Step 3

Multiply both sides by $15$ .

$\frac{y - 1}{15} \cdot 15 < - \frac{2}{3} \cdot 15$

Step 4

Simplify.

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Step 4.1

Simplify the left side.

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Step 4.1.1

Cancel the common factor of $15$ .

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Step 4.1.1.1

Cancel the common factor.

$\frac{y - 1}{15} \cdot 15 < - \frac{2}{3} \cdot 15$

Step 4.1.1.2

Rewrite the expression.

$y - 1 < - \frac{2}{3} \cdot 15$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $- \frac{2}{3} \cdot 15$ .

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Step 4.2.1.1

Cancel the common factor of $3$ .

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Step 4.2.1.1.1

Move the leading negative in $- \frac{2}{3}$ into the numerator.

$y - 1 < \frac{- 2}{3} \cdot 15$

Step 4.2.1.1.2

Factor $3$ out of $15$ .

$y - 1 < \frac{- 2}{3} \cdot (3 (5))$

Step 4.2.1.1.3

Cancel the common factor.

$y - 1 < \frac{- 2}{3} \cdot (3 \cdot 5)$

Step 4.2.1.1.4

Rewrite the expression.

$y - 1 < - 2 \cdot 5$

Step 4.2.1.2

Multiply $- 2$ by $5$ .

$y - 1 < - 10$

Step 5

Move all terms not containing $y$ to the right side of the inequality.

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Step 5.1

Add $1$ to both sides of the inequality.

$y < - 10 + 1$

Step 5.2

Add $- 10$ and $1$ .

$y < - 9$

Step 6

The result can be shown in multiple forms.

Inequality Form:

$y < - 9$

Interval Notation:

$(- \infty, - 9)$