Simplify -4<-5/(2y)+4/3

$- 4 < - \frac{5}{2 y} + \frac{4}{3}$

Step 1

Rewrite so $y$ is on the left side of the inequality.

$- \frac{5}{2 y} + \frac{4}{3} > - 4$

Step 2

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

Subtract $\frac{4}{3}$ from both sides of the inequality.

$- \frac{5}{2 y} > - 4 - \frac{4}{3}$

To write $- 4$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$- \frac{5}{2 y} > - 4 \cdot \frac{3}{3} - \frac{4}{3}$

Combine $- 4$ and $\frac{3}{3}$ .

$- \frac{5}{2 y} > \frac{- 4 \cdot 3}{3} - \frac{4}{3}$

Combine the numerators over the common denominator.

$- \frac{5}{2 y} > \frac{- 4 \cdot 3 - 4}{3}$

Simplify the numerator.

Multiply $- 4$ by $3$ .

$- \frac{5}{2 y} > \frac{- 12 - 4}{3}$

Subtract $4$ from $- 12$ .

$- \frac{5}{2 y} > \frac{- 16}{3}$

Move the negative in front of the fraction.

$- \frac{5}{2 y} > - \frac{16}{3}$

Step 3

Multiply both sides by $2 y$ .

$- \frac{5}{2 y} (2 y) = - \frac{16}{3} (2 y)$

Step 4

Simplify.

Simplify the left side.

Cancel the common factor of $2 y$ .

Move the leading negative in $- \frac{5}{2 y}$ into the numerator.

$\frac{- 5}{2 y} (2 y) = - \frac{16}{3} (2 y)$

Cancel the common factor.

$\frac{- 5}{2 y} (2 y) = - \frac{16}{3} (2 y)$

Rewrite the expression.

$- 5 = - \frac{16}{3} (2 y)$

Simplify the right side.

Simplify $- \frac{16}{3} (2 y)$ .

Multiply $- \frac{16}{3} (2 y)$ .

Multiply $2$ by $- 1$ .

$- 5 = - 2 (\frac{16}{3}) y$

Combine $- 2$ and $\frac{16}{3}$ .

$- 5 = \frac{- 2 \cdot 16}{3} y$

Multiply $- 2$ by $16$ .

$- 5 = \frac{- 32}{3} y$

Combine $\frac{- 32}{3}$ and $y$ .

$- 5 = \frac{- 32 y}{3}$

Move the negative in front of the fraction.

$- 5 = - \frac{32 y}{3}$

Step 5

Solve for $y$ .

Rewrite the equation as $- \frac{32 y}{3} = - 5$ .

$- \frac{32 y}{3} = - 5$

Multiply both sides of the equation by $- \frac{3}{32}$ .

$- \frac{3}{32} (- \frac{32 y}{3}) = - \frac{3}{32} \cdot - 5$

Simplify both sides of the equation.

Simplify the left side.

Simplify $- \frac{3}{32} (- \frac{32 y}{3})$ .

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{3}{32}$ into the numerator.

$\frac{- 3}{32} (- \frac{32 y}{3}) = - \frac{3}{32} \cdot - 5$

Move the leading negative in $- \frac{32 y}{3}$ into the numerator.

$\frac{- 3}{32} \cdot \frac{- 32 y}{3} = - \frac{3}{32} \cdot - 5$

Factor $3$ out of $- 3$ .

$\frac{3 (- 1)}{32} \cdot \frac{- 32 y}{3} = - \frac{3}{32} \cdot - 5$

Cancel the common factor.

$\frac{3 \cdot - 1}{32} \cdot \frac{- 32 y}{3} = - \frac{3}{32} \cdot - 5$

Rewrite the expression.

$\frac{- 1}{32} (- 32 y) = - \frac{3}{32} \cdot - 5$

Cancel the common factor of $32$ .

Factor $32$ out of $- 32 y$ .

$\frac{- 1}{32} (32 (- y)) = - \frac{3}{32} \cdot - 5$

Cancel the common factor.

$\frac{- 1}{32} (32 (- y)) = - \frac{3}{32} \cdot - 5$

Rewrite the expression.

$- - y = - \frac{3}{32} \cdot - 5$

Multiply.

Multiply $- 1$ by $- 1$ .

$1 y = - \frac{3}{32} \cdot - 5$

Multiply $y$ by $1$ .

$y = - \frac{3}{32} \cdot - 5$

Simplify the right side.

Multiply $- \frac{3}{32} \cdot - 5$ .

Multiply $- 5$ by $- 1$ .

$y = 5 (\frac{3}{32})$

Combine $5$ and $\frac{3}{32}$ .

$y = \frac{5 \cdot 3}{32}$

Multiply $5$ by $3$ .

$y = \frac{15}{32}$

Step 6

Find the domain of $\frac{5}{2 y} - \frac{16}{3}$ .

Set the denominator in $\frac{5}{2 y}$ equal to $0$ to find where the expression is undefined.

$2 y = 0$

Divide each term in $2 y = 0$ by $2$ and simplify.

Divide each term in $2 y = 0$ by $2$ .

$\frac{2 y}{2} = \frac{0}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 y}{2} = \frac{0}{2}$

Divide $y$ by $1$ .

$y = \frac{0}{2}$

Simplify the right side.

Divide $0$ by $2$ .

$y = 0$

The domain is all values of $y$ that make the expression defined.

$(- \infty, 0) \cup (0, \infty)$

Step 7

Use each root to create test intervals.

$y < 0$

$0 < y < \frac{15}{32}$

$y > \frac{15}{32}$

Step 8

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $y < 0$ to see if it makes the inequality true.

Choose a value on the interval $y < 0$ and see if this value makes the original inequality true.

$y = - 2$

Replace $y$ with $- 2$ in the original inequality.

$- 4 < - \frac{5}{2 (- 2)} + \frac{4}{3}$

The left side $- 4$ is less than the right side $2.58\overline{3}$ , which means that the given statement is always true.

True

Test a value on the interval $0 < y < \frac{15}{32}$ to see if it makes the inequality true.

Choose a value on the interval $0 < y < \frac{15}{32}$ and see if this value makes the original inequality true.

$y = 0.23$

Replace $y$ with $0.23$ in the original inequality.

$- 4 < - \frac{5}{2 (0.23)} + \frac{4}{3}$

The left side $- 4$ is not less than the right side $- 9.53623188$ , which means that the given statement is false.

False

Test a value on the interval $y > \frac{15}{32}$ to see if it makes the inequality true.

Choose a value on the interval $y > \frac{15}{32}$ and see if this value makes the original inequality true.

$y = 3$

Replace $y$ with $3$ in the original inequality.

$- 4 < - \frac{5}{2 (3)} + \frac{4}{3}$

The left side $- 4$ is less than the right side $0.5$ , which means that the given statement is always true.

True

Compare the intervals to determine which ones satisfy the original inequality.

$y < 0$ True

$0 < y < \frac{15}{32}$ False

$y > \frac{15}{32}$ True

$y < 0$ True

$0 < y < \frac{15}{32}$ False

$y > \frac{15}{32}$ True

Step 9

The solution consists of all of the true intervals.

$y < 0$ or $y > \frac{15}{32}$

Step 10

The result can be shown in multiple forms.

Inequality Form:

$y < 0 or y > \frac{15}{32}$

Interval Notation:

$(- \infty, 0) \cup (\frac{15}{32}, \infty)$

Step 11

Simplify -4<-5/(2y)+4/3

Related Questions

Support LCMGCF.com by disabling your adblocker.