# Chapter 4 - Proportions, Percents, and Solving Inequalities - 4.5 - Inequalities, Compound Inequalities, and Problem Solving - Problem Set 4.5 - Page 175: 43

$x\gt0$

#### Work Step by Step

Multiplying both sides by the $LCD= 10$ and then using the properties of inequalities, the solution to the given inequality, $\dfrac{x-1}{2}+\dfrac{x+3}{5}\gt\dfrac{1}{10} ,$ is \begin{array}{l}\require{cancel} 5(x-1)+2(x+3)\gt1(1) \\\\ 5x-5+2x+6\gt1 \\\\ 5x+2x\gt1+5-6 \\\\ 7x\gt0 \\\\ x\gt\dfrac{0}{7} \\\\ x\gt0 .\end{array}

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