Solve for x
x\in (-\infty,0)\cup [\frac{3}{10},\infty)
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\frac{3}{6x}+\frac{2\times 2x}{6x}\leq \frac{7}{3}
To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 6x and 3 is 6x. Multiply \frac{2}{3} times \frac{2x}{2x}.
\frac{3+2\times 2x}{6x}\leq \frac{7}{3}
Since \frac{3}{6x} and \frac{2\times 2x}{6x} have the same denominator, add them by adding their numerators.
\frac{3+4x}{6x}\leq \frac{7}{3}
Do the multiplications in 3+2\times 2x.
-\frac{1}{6}\times \frac{1}{x}\left(10x-3\right)\leq 0
Factor out x.
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