Solve x+3/4leq8x+2/3

$$x + \frac { 3 } { 4 } \leq \frac { 8 x + 2 } { 3 }$$

Solve for x

$$x\geq \frac{1}{20}$$

Solution Steps

Multiply both sides of the equation by 12, the least common multiple of 4,3. Since 12 is positive, the inequality direction remains the same.

$$12x+9\leq 4\left(8x+2\right)$$

Use the distributive property to multiply 4 by 8x+2.

$$12x+9\leq 32x+8$$

Subtract 32x from both sides.

$$12x+9-32x\leq 8$$

Combine 12x and -32x to get -20x.

$$-20x+9\leq 8$$

Subtract 9 from both sides.

$$-20x\leq 8-9$$

Subtract 9 from 8 to get -1.

$$-20x\leq -1$$

Divide both sides by -20. Since -20 is negative, the inequality direction is changed.

$$x\geq \frac{-1}{-20}$$

Fraction \frac{-1}{-20} can be simplified to \frac{1}{20} by removing the negative sign from both the numerator and the denominator.

$$x\geq \frac{1}{20}$$