Solve the inequality.Express the solution set in interval notation. $$0

Question

Solve the inequality.Express the solution set in interval notation. $$0 < | 8 x + 3 | < 11$$
A) $\quad \left( - \frac { 7 } { 4 } , - \frac { 3 } { 8 } \right) \cup \left( - \frac { 3 } { 8 } , 1 \right)$
B) $\quad \left( - \frac { 3 } { 8 } , 1 \right)$
C) $\quad \left( - \frac { 7 } { 4 } , - \frac { 3 } { 8 } \right)$
D) $\quad ( - \infty , - 1 ) \cup \left( \frac { 3 } { 8 } , \infty \right)$
E) $\left( - \infty , - \frac { 3 } { 8 } \right) \cup ( 1 , \infty )$

Quizplus · Accepted Answer

To solve the inequality $0 < | 8x + 3 | < 11$, we break it into two parts: $| 8x + 3 | > 0$ and $| 8x + 3 | < 11$.1. For $| 8x + 3 | > 0$, any real number except $x = -\frac{3}{8}$ will satisfy this condition, as absolute value is always non-negative and is zero only when the expression inside is zero.2. For $| 8x + 3 | < 11$, we solve it as two separate inequalities because the absolute value being less than 11 means the expression inside can be between -11 and 11 (not including -11 and 11 themselves). a. $8x + 3 < 11$ simplifies to $8x < 8$, which gives $x < 1$. b. $8x + 3 > -11$ simplifies to $8x > -14$, which gives $x > -\frac{7}{4}$.Combining these, we get $x > -\frac{7}{4}$ and $x < 1$, but $x$ cannot be $-\frac{3}{8}$ because of the first condition. Therefore, the solution set in interval notation is $\left( - \frac { 7 } { 4 } , - \frac { 3 } { 8 } \right) \cup \left( - \frac { 3 } { 8 } , 1 \right)$.

Solve the Inequality 0<∣8x+3∣<110 < | 8 x + 3 | < 110<∣8x+3∣<11

Solve the Inequality $0 < | 8 x + 3 | < 11$