Convert to Set Notation -3/4x>=-7/12+2/3x

$- \frac{3}{4} x \geq - \frac{7}{12} + \frac{2}{3} x$

Step 1

Solve $x \leq \frac{7}{17}$ .

Simplify $- \frac{3}{4} x$ .

Rewrite.

$0 + 0 - \frac{3}{4} x \geq - \frac{7}{12} + \frac{2}{3} x$

Simplify by adding zeros.

$- \frac{3}{4} x \geq - \frac{7}{12} + \frac{2}{3} x$

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

$- \frac{x \cdot 3}{4} \geq - \frac{7}{12} + \frac{2}{3} x$

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

$- \frac{3 x}{4} \geq - \frac{7}{12} + \frac{2}{3} x$

Combine $\frac{2}{3}$ and $x$ .

$- \frac{3 x}{4} \geq - \frac{7}{12} + \frac{2 x}{3}$

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

Subtract $\frac{2 x}{3}$ from both sides of the inequality.

$- \frac{3 x}{4} - \frac{2 x}{3} \geq - \frac{7}{12}$

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

$- \frac{3 x}{4} \cdot \frac{3}{3} - \frac{2 x}{3} \geq - \frac{7}{12}$

To write $- \frac{2 x}{3}$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$- \frac{3 x}{4} \cdot \frac{3}{3} - \frac{2 x}{3} \cdot \frac{4}{4} \geq - \frac{7}{12}$

Write each expression with a common denominator of $12$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{3 x}{4}$ by $\frac{3}{3}$ .

$- \frac{3 x \cdot 3}{4 \cdot 3} - \frac{2 x}{3} \cdot \frac{4}{4} \geq - \frac{7}{12}$

Multiply $4$ by $3$ .

$- \frac{3 x \cdot 3}{12} - \frac{2 x}{3} \cdot \frac{4}{4} \geq - \frac{7}{12}$

Multiply $\frac{2 x}{3}$ by $\frac{4}{4}$ .

$- \frac{3 x \cdot 3}{12} - \frac{2 x \cdot 4}{3 \cdot 4} \geq - \frac{7}{12}$

Multiply $3$ by $4$ .

$- \frac{3 x \cdot 3}{12} - \frac{2 x \cdot 4}{12} \geq - \frac{7}{12}$

Combine the numerators over the common denominator.

$\frac{- 3 x \cdot 3 - 2 x \cdot 4}{12} \geq - \frac{7}{12}$

Simplify the numerator.

Factor $x$ out of $- 3 x \cdot 3 - 2 x \cdot 4$ .

Factor $x$ out of $- 3 x \cdot 3$ .

$\frac{x (- 3 \cdot 3) - 2 x \cdot 4}{12} \geq - \frac{7}{12}$

Factor $x$ out of $- 2 x \cdot 4$ .

$\frac{x (- 3 \cdot 3) + x (- 2 \cdot 4)}{12} \geq - \frac{7}{12}$

Factor $x$ out of $x (- 3 \cdot 3) + x (- 2 \cdot 4)$ .

$\frac{x (- 3 \cdot 3 - 2 \cdot 4)}{12} \geq - \frac{7}{12}$

Multiply $- 3$ by $3$ .

$\frac{x (- 9 - 2 \cdot 4)}{12} \geq - \frac{7}{12}$

Multiply $- 2$ by $4$ .

$\frac{x (- 9 - 8)}{12} \geq - \frac{7}{12}$

Subtract $8$ from $- 9$ .

$\frac{x \cdot - 17}{12} \geq - \frac{7}{12}$

Move $- 17$ to the left of $x$ .

$\frac{- 17 \cdot x}{12} \geq - \frac{7}{12}$

Move the negative in front of the fraction.

$- \frac{17 x}{12} \geq - \frac{7}{12}$

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$- 17 x \geq - 7$

Divide each term in $- 17 x \geq - 7$ by $- 17$ and simplify.

Divide each term in $- 17 x \geq - 7$ by $- 17$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 17 x}{- 17} \leq \frac{- 7}{- 17}$

Simplify the left side.

Cancel the common factor of $- 17$ .

Cancel the common factor.

$\frac{- 17 x}{- 17} \leq \frac{- 7}{- 17}$

Divide $x$ by $1$ .

$x \leq \frac{- 7}{- 17}$

Simplify the right side.

Dividing two negative values results in a positive value.

$x \leq \frac{7}{17}$

Step 2

Use the inequality $x \leq \frac{7}{17}$ to build the set notation.

${x | x \leq \frac{7}{17}}$

Step 3

Convert to Set Notation -3/4x>=-7/12+2/3x

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