Solve the inequality. $$- \frac { 7 } { 5 } - 4 - \frac { 1 } { 2 } y \text { or } \frac { 7 } { 6 } 2 y + \frac { 4 } { 3 }$$
A){ }
B) $( - \infty , \infty )$
C) $\left( - \infty , - \frac { 1 } { 12 } \right)$
D) $\left( - \frac { 26 } { 5 } , - \frac { 1 } { 12 } \right)$

Question

Solve the inequality. $$- \frac { 7 } { 5 } > - 4 - \frac { 1 } { 2 } y \text { or } \frac { 7 } { 6 } > 2 y + \frac { 4 } { 3 }$$
A){  }
B) $( - \infty , \infty )$
C) $\left( - \infty , - \frac { 1 } { 12 } \right)$
D) $\left( - \frac { 26 } { 5 } , - \frac { 1 } { 12 } \right)$

Quizplus · Accepted Answer

To solve the given inequality, we break it into two parts and solve each separately:1. $- \frac { 7 } { 5 } > - 4 - \frac { 1 } { 2 } y$Rearranging, we get: $\frac { 1 } { 2 } y > - 4 + \frac { 7 } { 5 }$Simplifying further: $y > - \frac { 20 } { 5 } + \frac { 7 } { 5 }$$y > - \frac { 13 } { 5 }$2. $\frac { 7 } { 6 } > 2 y + \frac { 4 } { 3 }$Rearranging, we get: $2 y < \frac { 7 } { 6 } - \frac { 4 } { 3 }$Simplifying further: $y < - \frac { 1 } { 12 }$Combining both parts, we find that all values of $y$ satisfy at least one part of the inequality, meaning the solution is all real numbers, or $( - \infty , \infty )$.

Solve the Inequality −75>−4−12y or 76>2y+43- \frac { 7 } { 5 } > - 4 - \frac { 1 } { 2 } y \text { or } \frac { 7 } { 6 } > 2 y + \frac { 4 } { 3 }−57​>−4−21​y or 67​>2y+34​

Solve the Inequality $- \frac { 7 } { 5 } > - 4 - \frac { 1 } { 2 } y \text { or } \frac { 7 } { 6 } > 2 y + \frac { 4 } { 3 }$