Solve the Inequality for y 8+3/y>19/y

$8 + \frac{3}{y} > \frac{19}{y}$

Step 1

Multiply both sides by $y$ .

$(8 + \frac{3}{y}) y = \frac{19}{y} y$

Step 2

Simplify.

Simplify the left side.

Simplify $(8 + \frac{3}{y}) y$ .

Apply the distributive property.

$8 y + \frac{3}{y} y = \frac{19}{y} y$

Cancel the common factor of $y$ .

Cancel the common factor.

$8 y + \frac{3}{y} y = \frac{19}{y} y$

Rewrite the expression.

$8 y + 3 = \frac{19}{y} y$

Simplify the right side.

Cancel the common factor of $y$ .

Cancel the common factor.

$8 y + 3 = \frac{19}{y} y$

Rewrite the expression.

$8 y + 3 = 19$

Step 3

Solve for $y$ .

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

Subtract $3$ from both sides of the equation.

$8 y = 19 - 3$

Subtract $3$ from $19$ .

$8 y = 16$

Divide each term in $8 y = 16$ by $8$ and simplify.

Divide each term in $8 y = 16$ by $8$ .

$\frac{8 y}{8} = \frac{16}{8}$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 y}{8} = \frac{16}{8}$

Divide $y$ by $1$ .

$y = \frac{16}{8}$

Simplify the right side.

Divide $16$ by $8$ .

$y = 2$

Step 4

Find the domain of $8 - \frac{16}{y}$ .

Set the denominator in $\frac{16}{y}$ equal to $0$ to find where the expression is undefined.

$y = 0$

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

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

Step 5

Use each root to create test intervals.

$y < 0$

$0 < y < 2$

$y > 2$

Step 6

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.

$8 + \frac{3}{- 2} > \frac{19}{- 2}$

The left side $6.5$ is greater than the right side $- 9.5$ , which means that the given statement is always true.

True

Test a value on the interval $0 < y < 2$ to see if it makes the inequality true.

Choose a value on the interval $0 < y < 2$ and see if this value makes the original inequality true.

$y = 1$

Replace $y$ with $1$ in the original inequality.

$8 + \frac{3}{1} > \frac{19}{1}$

The left side $11$ is not greater than the right side $19$ , which means that the given statement is false.

False

Test a value on the interval $y > 2$ to see if it makes the inequality true.

Choose a value on the interval $y > 2$ and see if this value makes the original inequality true.

$y = 4$

Replace $y$ with $4$ in the original inequality.

$8 + \frac{3}{4} > \frac{19}{4}$

The left side $8.75$ is greater than the right side $4.75$ , which means that the given statement is always true.

True

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

$y < 0$ True

$0 < y < 2$ False

$y > 2$ True

$y < 0$ True

$0 < y < 2$ False

$y > 2$ True

Step 7

The solution consists of all of the true intervals.

$y < 0$ or $y > 2$

Step 8

The result can be shown in multiple forms.

Inequality Form:

$y < 0 or y > 2$

Interval Notation:

$(- \infty, 0) \cup (2, \infty)$

Step 9

Solve the Inequality for y 8+3/y>19/y

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