Solve for d (d-3)/10>=(2d+3)/5+(d+3)/3

$\frac{d - 3}{10} \geq \frac{2 d + 3}{5} + \frac{d + 3}{3}$

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

$\frac{2 d + 3}{5} + \frac{d + 3}{3} \leq \frac{d - 3}{10}$

Multiply both sides by $10$ .

$(\frac{2 d + 3}{5} + \frac{d + 3}{3}) \cdot 10 \leq \frac{d - 3}{10} \cdot 10$

Simplify.

Simplify the left side.

Simplify $(\frac{2 d + 3}{5} + \frac{d + 3}{3}) \cdot 10$ .

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

$(\frac{2 d + 3}{5} \cdot \frac{3}{3} + \frac{d + 3}{3}) \cdot 10 \leq \frac{d - 3}{10} \cdot 10$

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

$(\frac{2 d + 3}{5} \cdot \frac{3}{3} + \frac{d + 3}{3} \cdot \frac{5}{5}) \cdot 10 \leq \frac{d - 3}{10} \cdot 10$

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

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

$(\frac{(2 d + 3) \cdot 3}{5 \cdot 3} + \frac{d + 3}{3} \cdot \frac{5}{5}) \cdot 10 \leq \frac{d - 3}{10} \cdot 10$

Multiply $5$ by $3$ .

$(\frac{(2 d + 3) \cdot 3}{15} + \frac{d + 3}{3} \cdot \frac{5}{5}) \cdot 10 \leq \frac{d - 3}{10} \cdot 10$

Multiply $\frac{d + 3}{3}$ by $\frac{5}{5}$ .

$(\frac{(2 d + 3) \cdot 3}{15} + \frac{(d + 3) \cdot 5}{3 \cdot 5}) \cdot 10 \leq \frac{d - 3}{10} \cdot 10$

Multiply $3$ by $5$ .

$(\frac{(2 d + 3) \cdot 3}{15} + \frac{(d + 3) \cdot 5}{15}) \cdot 10 \leq \frac{d - 3}{10} \cdot 10$

Combine the numerators over the common denominator.

$\frac{(2 d + 3) \cdot 3 + (d + 3) \cdot 5}{15} \cdot 10 \leq \frac{d - 3}{10} \cdot 10$

Simplify the numerator.

Apply the distributive property.

$\frac{2 d \cdot 3 + 3 \cdot 3 + (d + 3) \cdot 5}{15} \cdot 10 \leq \frac{d - 3}{10} \cdot 10$

Multiply $3$ by $2$ .

$\frac{6 d + 3 \cdot 3 + (d + 3) \cdot 5}{15} \cdot 10 \leq \frac{d - 3}{10} \cdot 10$

Multiply $3$ by $3$ .

$\frac{6 d + 9 + (d + 3) \cdot 5}{15} \cdot 10 \leq \frac{d - 3}{10} \cdot 10$

Apply the distributive property.

$\frac{6 d + 9 + d \cdot 5 + 3 \cdot 5}{15} \cdot 10 \leq \frac{d - 3}{10} \cdot 10$

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

$\frac{6 d + 9 + 5 \cdot d + 3 \cdot 5}{15} \cdot 10 \leq \frac{d - 3}{10} \cdot 10$

Multiply $3$ by $5$ .

$\frac{6 d + 9 + 5 \cdot d + 15}{15} \cdot 10 \leq \frac{d - 3}{10} \cdot 10$

Add $6 d$ and $5 d$ .

$\frac{11 d + 9 + 15}{15} \cdot 10 \leq \frac{d - 3}{10} \cdot 10$

Add $9$ and $15$ .

$\frac{11 d + 24}{15} \cdot 10 \leq \frac{d - 3}{10} \cdot 10$

Simplify terms.

Cancel the common factor of $5$ .

Factor $5$ out of $15$ .

$\frac{11 d + 24}{5 (3)} \cdot 10 \leq \frac{d - 3}{10} \cdot 10$

Factor $5$ out of $10$ .

$\frac{11 d + 24}{5 \cdot 3} \cdot (5 \cdot 2) \leq \frac{d - 3}{10} \cdot 10$

Cancel the common factor.

$\frac{11 d + 24}{5 \cdot 3} \cdot (5 \cdot 2) \leq \frac{d - 3}{10} \cdot 10$

Rewrite the expression.

$\frac{11 d + 24}{3} \cdot 2 \leq \frac{d - 3}{10} \cdot 10$

Combine $\frac{11 d + 24}{3}$ and $2$ .

$\frac{(11 d + 24) \cdot 2}{3} \leq \frac{d - 3}{10} \cdot 10$

Move $2$ to the left of $11 d + 24$ .

$\frac{2 (11 d + 24)}{3} \leq \frac{d - 3}{10} \cdot 10$

Simplify the right side.

Cancel the common factor of $10$ .

Cancel the common factor.

$\frac{2 (11 d + 24)}{3} \leq \frac{d - 3}{10} \cdot 10$

Rewrite the expression.

$\frac{2 (11 d + 24)}{3} \leq d - 3$

Solve for $d$ .

Multiply both sides by $3$ .

$\frac{2 (11 d + 24)}{3} \cdot 3 \leq (d - 3) \cdot 3$

Simplify.

Simplify the left side.

Simplify $\frac{2 (11 d + 24)}{3} \cdot 3$ .

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{2 (11 d + 24)}{3} \cdot 3 \leq (d - 3) \cdot 3$

Rewrite the expression.

$2 (11 d + 24) \leq (d - 3) \cdot 3$

Apply the distributive property.

$2 (11 d) + 2 \cdot 24 \leq (d - 3) \cdot 3$

Multiply.

Multiply $11$ by $2$ .

$22 d + 2 \cdot 24 \leq (d - 3) \cdot 3$

Multiply $2$ by $24$ .

$22 d + 48 \leq (d - 3) \cdot 3$

Simplify the right side.

Simplify $(d - 3) \cdot 3$ .

Apply the distributive property.

$22 d + 48 \leq d \cdot 3 - 3 \cdot 3$

Simplify the expression.

Move $3$ to the left of $d$ .

$22 d + 48 \leq 3 \cdot d - 3 \cdot 3$

Multiply $- 3$ by $3$ .

$22 d + 48 \leq 3 d - 9$

Solve for $d$ .

Move all terms containing $d$ to the left side of the inequality.

Subtract $3 d$ from both sides of the inequality.

$22 d + 48 - 3 d \leq - 9$

Subtract $3 d$ from $22 d$ .

$19 d + 48 \leq - 9$

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

Subtract $48$ from both sides of the inequality.

$19 d \leq - 9 - 48$

Subtract $48$ from $- 9$ .

$19 d \leq - 57$

Divide each term in $19 d \leq - 57$ by $19$ and simplify.

Divide each term in $19 d \leq - 57$ by $19$ .

$\frac{19 d}{19} \leq \frac{- 57}{19}$

Simplify the left side.

Cancel the common factor of $19$ .

Cancel the common factor.

$\frac{19 d}{19} \leq \frac{- 57}{19}$

Divide $d$ by $1$ .

$d \leq \frac{- 57}{19}$

Simplify the right side.

Divide $- 57$ by $19$ .

$d \leq - 3$

The result can be shown in multiple forms.

Inequality Form:

$d \leq - 3$

Interval Notation:

$(- \infty, - 3]$

Solve for d (d-3)/10>=(2d+3)/5+(d+3)/3

Related Questions

Support LCMGCF.com by disabling your adblocker.