Solve each compound inequality. Express the solution sets in interval notation. Match each compound inequality with the interval to obtain a true statement.
- $( - \infty , - 6 ) \cup \left( - \frac { 1 } { 13 } , \infty \right)$

Question

Solve each compound inequality. Express the solution sets in interval notation. Match each compound inequality with the interval to obtain a true statement. -$( - \infty , - 6 ) \cup \left( - \frac { 1 } { 13 } , \infty \right)$ A) $15 x - 14 < 0 \text { and } 13 x - 1 > 0$ B) $x + 1 > 0 \text { and } 13 x - 14 < 0$ C) $13 x + 14 < - 64 \text { or } 13 x + 14 > 13$ D) $15 x - 14 < - 14 \text { or } 15 x - 8 > 8$

Quizplus · Accepted Answer

To solve this compound inequality, we need to find the values of x for which either of the inequalities is true. For the first inequality: $15x-14<0$ $15x<14$ $x<\frac{14}{15}$ For the second inequality: $13x-1>0$ $13x>1$ $x>\frac{1}{13}$ Putting the values on the line, we get: <----(x>$\frac{1}{13}$)---------------($\frac{14}{15}$ This means that any value of x which is either less than $\frac{1}{13}$ or greater than $\frac{14}{15}$ will satisfy the given compound inequality. The interval notation for this solution set is: $(-\infty,-\frac{1}{13}) \cup (\frac{14}{15},\infty)$ Matching this with the options, we see that option C has the same solution set as the given compound inequality.

Solve Each Compound Inequality (−∞,−6)∪(−113,∞)( - \infty , - 6 ) \cup \left( - \frac { 1 } { 13 } , \infty \right)(−∞,−6)∪(−131​,∞)

Solve Each Compound Inequality $( - \infty , - 6 ) \cup \left( - \frac { 1 } { 13 } , \infty \right)$