Solve -1/2x-1/3leq6x+2

$$- \frac{ 1 }{ 2 } x- \frac{ 1 }{ 3 } \leq 6x+2$$

Solve for x

$$x\geq -\frac{14}{39}$$

Solution Steps

Subtract 6x from both sides.

$$-\frac{1}{2}x-\frac{1}{3}-6x\leq 2$$

Combine -\frac{1}{2}x and -6x to get -\frac{13}{2}x.

$$-\frac{13}{2}x-\frac{1}{3}\leq 2$$

Add \frac{1}{3} to both sides.

$$-\frac{13}{2}x\leq 2+\frac{1}{3}$$

Convert 2 to fraction \frac{6}{3}.

$$-\frac{13}{2}x\leq \frac{6}{3}+\frac{1}{3}$$

Since \frac{6}{3} and \frac{1}{3} have the same denominator, add them by adding their numerators.

$$-\frac{13}{2}x\leq \frac{6+1}{3}$$

Add 6 and 1 to get 7.

$$-\frac{13}{2}x\leq \frac{7}{3}$$

Multiply both sides by -\frac{2}{13}, the reciprocal of -\frac{13}{2}. Since -\frac{13}{2} is negative, the inequality direction is changed.

$$x\geq \frac{7}{3}\left(-\frac{2}{13}\right)$$

Multiply \frac{7}{3} times -\frac{2}{13} by multiplying numerator times numerator and denominator times denominator.

$$x\geq \frac{7\left(-2\right)}{3\times 13}$$

Do the multiplications in the fraction \frac{7\left(-2\right)}{3\times 13}.

$$x\geq \frac{-14}{39}$$

Fraction \frac{-14}{39} can be rewritten as -\frac{14}{39} by extracting the negative sign.

$$x\geq -\frac{14}{39}$$