Solve the inequality x−5<2x5≤x2+15\displaystyle x - 5 < \frac{2x}{5} \le \frac{x}{2} + \frac{1}{5}. Answer ...

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Worked solution

6

1. Separate the compound inequality into two inequalities.

x−5<2x5and2x5⩽x2+15x-5<\frac{2x}{5}\quad \text{and}\quad \frac{2x}{5}\leqslant \frac{x}{2}+\frac{1}{5}

2. Solve the left inequality to find the upper bound for $x$.

x−5<2x5x-5<\frac{2x}{5}
Multiply by 55:
5x−25<2x5x-25<2x
3x<253x<25
x<253x<\frac{25}{3}

3. Solve the right inequality to find the lower bound for $x$.

2x5⩽x2+15\frac{2x}{5}\leqslant \frac{x}{2}+\frac{1}{5}
Multiply by 1010:
4x⩽5x+24x\leqslant 5x+2
−x⩽2-x\leqslant 2
Divide by −1-1, reversing the inequality sign:
x⩾−2x\geqslant -2

4. Combine the two solution ranges.

Since both inequalities must be true,
x⩾−2andx<253x\geqslant -2 \quad \text{and}\quad x<\frac{25}{3}
Hence,
−2⩽x<253-2\leqslant x<\frac{25}{3}

Answer

−4⩽x<253-4 \leqslant x < \dfrac{25}{3}

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