How do you solve $\frac{1}{2} x - \frac{1}{10} \leq x + 1$ $1/2 x - 1/10<=x + 1$ ?

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How do you solve $\frac{1}{2} x - \frac{1}{10} \leq x + 1$ $1/2 x - 1/10<=x + 1$ ?

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Answer 1 · 2016-01-08T23:37:07

x $\geq$ $>=$ -2.2

Explanation:

If one subtracts x from both sides of the inequality, one gets $- \frac{1}{2} x - \frac{1}{10} \leq 1$ $-1/2x -1/10 <= 1$ . So, by adding $\frac{1}{10}$ $1/10$ to both sides, one gets $- \frac{1}{2} x \leq 1.1$ $-1/2 x <= 1.1$ . Multiplying both sides by 2, one gets $- x \geq 2.2$ $-x >= 2.2$ . Multiplying both sides of the inequality by -1, which reverses the inequality from $\leq$ $<=$ to $\geq$ $>=$ , one gets $x \geq - 2.2$ $x >= -2.2$ .

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How do you solve $\frac{1}{2} x - \frac{1}{10} \leq x + 1$ $1/2 x - 1/10<=x + 1$ ?

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