Answer
The provided statement is false and the correct statement is “The inequalities $\left( x+3 \right)\left( x-1 \right)\ge 0$ and $\frac{x+3}{x-1}\ge 0$ have different solution sets.
Work Step by Step
First consider the inequality $\left( x+3 \right)\left( x-1 \right)\ge 0$.
Solve this to get, $x\le -3$ and $x\ge 1$
Hence, the solution set of $\left( x+3 \right)\left( x-1 \right)\ge 0$ is $\left( -\infty ,-3 \right]\cup \left[ 1,\infty \right)$.
Now, consider the inequality $\frac{x+3}{x-1}\ge 0$.
Divide both sides by ${{\left( x-1 \right)}^{2}},x\ne 1$.
${{\left( x-1 \right)}^{2}}\frac{x+3}{x-1}\ge 0$
$x\le -3$ And $x>1$
Hence, the solution set of $\frac{x+3}{x-1}\ge 0$ is $\left( -\infty ,-3 \right]\cup \left( 1,\infty \right)$.
Hence, the provided statement is false.
The provided statement is false and the correct statement is “The inequalities $\left( x+3 \right)\left( x-1 \right)\ge 0$ and $\frac{x+3}{x-1}\ge 0$ have different solution sets”.
