Solve the Inequality for x (x+3)/8-(x+5)/5>=3/10

$\frac{x + 3}{8} - \frac{x + 5}{5} \geq \frac{3}{10}$

Step 1

Simplify $\frac{x + 3}{8} - \frac{x + 5}{5}$ .

To write $\frac{x + 3}{8}$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$\frac{x + 3}{8} \cdot \frac{5}{5} - \frac{x + 5}{5} \geq \frac{3}{10}$

To write $- \frac{x + 5}{5}$ as a fraction with a common denominator, multiply by $\frac{8}{8}$ .

$\frac{x + 3}{8} \cdot \frac{5}{5} - \frac{x + 5}{5} \cdot \frac{8}{8} \geq \frac{3}{10}$

Write each expression with a common denominator of $40$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{x + 3}{8}$ by $\frac{5}{5}$ .

$\frac{(x + 3) \cdot 5}{8 \cdot 5} - \frac{x + 5}{5} \cdot \frac{8}{8} \geq \frac{3}{10}$

Multiply $8$ by $5$ .

$\frac{(x + 3) \cdot 5}{40} - \frac{x + 5}{5} \cdot \frac{8}{8} \geq \frac{3}{10}$

Multiply $\frac{x + 5}{5}$ by $\frac{8}{8}$ .

$\frac{(x + 3) \cdot 5}{40} - \frac{(x + 5) \cdot 8}{5 \cdot 8} \geq \frac{3}{10}$

Multiply $5$ by $8$ .

$\frac{(x + 3) \cdot 5}{40} - \frac{(x + 5) \cdot 8}{40} \geq \frac{3}{10}$

Combine the numerators over the common denominator.

$\frac{(x + 3) \cdot 5 - (x + 5) \cdot 8}{40} \geq \frac{3}{10}$

Simplify the numerator.

Apply the distributive property.

$\frac{x \cdot 5 + 3 \cdot 5 - (x + 5) \cdot 8}{40} \geq \frac{3}{10}$

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

$\frac{5 \cdot x + 3 \cdot 5 - (x + 5) \cdot 8}{40} \geq \frac{3}{10}$

Multiply $3$ by $5$ .

$\frac{5 \cdot x + 15 - (x + 5) \cdot 8}{40} \geq \frac{3}{10}$

Apply the distributive property.

$\frac{5 x + 15 + (- x - 1 \cdot 5) \cdot 8}{40} \geq \frac{3}{10}$

Multiply $- 1$ by $5$ .

$\frac{5 x + 15 + (- x - 5) \cdot 8}{40} \geq \frac{3}{10}$

Apply the distributive property.

$\frac{5 x + 15 - x \cdot 8 - 5 \cdot 8}{40} \geq \frac{3}{10}$

Multiply $8$ by $- 1$ .

$\frac{5 x + 15 - 8 x - 5 \cdot 8}{40} \geq \frac{3}{10}$

Multiply $- 5$ by $8$ .

$\frac{5 x + 15 - 8 x - 40}{40} \geq \frac{3}{10}$

Subtract $8 x$ from $5 x$ .

$\frac{- 3 x + 15 - 40}{40} \geq \frac{3}{10}$

Subtract $40$ from $15$ .

$\frac{- 3 x - 25}{40} \geq \frac{3}{10}$

Simplify with factoring out.

Factor $- 1$ out of $- 3 x$ .

$\frac{- (3 x) - 25}{40} \geq \frac{3}{10}$

Rewrite $- 25$ as $- 1 (25)$ .

$\frac{- (3 x) - 1 (25)}{40} \geq \frac{3}{10}$

Factor $- 1$ out of $- (3 x) - 1 (25)$ .

$\frac{- (3 x + 25)}{40} \geq \frac{3}{10}$

Simplify the expression.

Rewrite $- (3 x + 25)$ as $- 1 (3 x + 25)$ .

$\frac{- 1 (3 x + 25)}{40} \geq \frac{3}{10}$

Move the negative in front of the fraction.

$- \frac{3 x + 25}{40} \geq \frac{3}{10}$

Step 2

Divide each term in $- \frac{3 x + 25}{40} \geq \frac{3}{10}$ by $- 1$ and simplify.

Divide each term in $- \frac{3 x + 25}{40} \geq \frac{3}{10}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- \frac{3 x + 25}{40}}{- 1} \leq \frac{\frac{3}{10}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{\frac{3 x + 25}{40}}{1} \leq \frac{\frac{3}{10}}{- 1}$

Divide $\frac{3 x + 25}{40}$ by $1$ .

$\frac{3 x + 25}{40} \leq \frac{\frac{3}{10}}{- 1}$

Simplify the right side.

Move the negative one from the denominator of $\frac{\frac{3}{10}}{- 1}$ .

$\frac{3 x + 25}{40} \leq - 1 \cdot \frac{3}{10}$

Rewrite $- 1 \cdot \frac{3}{10}$ as $- \frac{3}{10}$ .

$\frac{3 x + 25}{40} \leq - \frac{3}{10}$

Step 3

Multiply both sides by $40$ .

$\frac{3 x + 25}{40} \cdot 40 \leq - \frac{3}{10} \cdot 40$

Step 4

Simplify.

Simplify the left side.

Cancel the common factor of $40$ .

Cancel the common factor.

$\frac{3 x + 25}{40} \cdot 40 \leq - \frac{3}{10} \cdot 40$

Rewrite the expression.

$3 x + 25 \leq - \frac{3}{10} \cdot 40$

Simplify the right side.

Simplify $- \frac{3}{10} \cdot 40$ .

Cancel the common factor of $10$ .

Move the leading negative in $- \frac{3}{10}$ into the numerator.

$3 x + 25 \leq \frac{- 3}{10} \cdot 40$

Factor $10$ out of $40$ .

$3 x + 25 \leq \frac{- 3}{10} \cdot (10 (4))$

Cancel the common factor.

$3 x + 25 \leq \frac{- 3}{10} \cdot (10 \cdot 4)$

Rewrite the expression.

$3 x + 25 \leq - 3 \cdot 4$

Multiply $- 3$ by $4$ .

$3 x + 25 \leq - 12$

Step 5

Solve for $x$ .

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

Subtract $25$ from both sides of the inequality.

$3 x \leq - 12 - 25$

Subtract $25$ from $- 12$ .

$3 x \leq - 37$

Divide each term in $3 x \leq - 37$ by $3$ and simplify.

Divide each term in $3 x \leq - 37$ by $3$ .

$\frac{3 x}{3} \leq \frac{- 37}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} \leq \frac{- 37}{3}$

Divide $x$ by $1$ .

$x \leq \frac{- 37}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x \leq - \frac{37}{3}$

Step 6

The result can be shown in multiple forms.

Inequality Form:

$x \leq - \frac{37}{3}$

Interval Notation:

$(- \infty, - \frac{37}{3}]$

Solve the Inequality for x (x+3)/8-(x+5)/5>=3/10

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