Solve for \( x \) (1) \( \frac{3}{x-1}>0 \) 2) \( \frac{-2}{x+3} \geq 0 \)
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Bonus Knowledge
To solve the inequality \( \frac{3}{x-1}>0 \), we need to find where the fraction is positive. Since the numerator \( 3 \) is always positive, we can focus on the denominator. The fraction is positive when \( x - 1 > 0 \), so \( x > 1 \). For the second inequality \( \frac{-2}{x+3} \geq 0 \), the numerator \( -2 \) is negative, so the fraction will be non-negative when the denominator \( x + 3 \) is negative, which means \( x + 3 < 0 \) or \( x < -3 \). Combining both results: 1. From the first inequality, \( x > 1 \). 2. From the second inequality, \( x < -3 \). Since there are no \( x \) values that can satisfy both conditions simultaneously, the solution set is empty. So, the final answer is that there are no solutions for \( x \). For a little historical touch, inequalities have been a staple of mathematics since ancient times, with the Greeks first exploring the idea of comparing quantities. Fast forward to today, inequalities help us make decisions in countless fields, from economics to engineering! If you’re wrestling with inequalities, a common mistake is forgetting to flip the inequality sign when multiplying or dividing by a negative number! Always double-check that detail to avoid sneaky errors.