Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 8 - Section 8.4 - Nonlinear Inequalities in One Variable - Exercise Set - Page 510: 21

Answer

$\color{blue}{(-\infty, 2) \cup (2.75, +\infty)}$

Work Step by Step

Find the value/s of $x$ that will make the denominator equal to zero by equating the denominator to zero then solving the equation. $$x-2=0 \longrightarrow x=2$$ Solve the related equation: $$\dfrac{3}{x-2}=4$$ Multiply $x-2$ to both sides of the equation to obtain: $$\\3=4(x-2) \\3=4x-8 \\11=4x \\\frac{11}{4}=x \\2.75=x$$ The numbers $2$ and $2.75$ divide the number line into three regions, namely: A $(-\infty, 2)$ B $(2, 2.75)$ C $(2.75, +\infty)$ Pick a test point value in each region and substitute it into the given inequality. If the test point satisfies the given inequality, then the region where the test point value belongs is a solution. (Refer to the table below.) The table below shows that the solution includes the intervals $(-\infty, 2)$ and $(2.75, +\infty)$. $2.75$ is not included because the inequality involved is $\lt$. $2$ is not included because it makes the denominator equal to zero. Therefore, the solution is: $\color{blue}{(-\infty, 2) \cup (2.75, +\infty)}$
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