Solve for x
x\leq -\frac{19}{8}
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-\frac{3}{8}\geq 2+x
Divide both sides by 8. Since 8 is positive, the inequality direction remains the same.
2+x\leq -\frac{3}{8}
Swap sides so that all variable terms are on the left hand side. This changes the sign direction.
x\leq -\frac{3}{8}-2
Subtract 2 from both sides.
x\leq -\frac{3}{8}-\frac{16}{8}
Convert 2 to fraction \frac{16}{8}.
x\leq \frac{-3-16}{8}
Since -\frac{3}{8} and \frac{16}{8} have the same denominator, subtract them by subtracting their numerators.
x\leq -\frac{19}{8}
Subtract 16 from -3 to get -19.
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Limits
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