Solve the inequality. Write the solution set in interval notation.
-$\frac { 9 h } { h - 1 } - 7$
A) $\left( \frac { 7 } { 16 } , \infty \right)$
B) $\left( \frac { 7 } { 16 } , 1 \right)$
C) $\left( - \infty , - \frac { 7 } { 2 } \right) \cup ( 1 , \infty )$
D) $\left( - \infty , \frac { 7 } { 16 } \right) \cup ( 1 , \infty )$

Question

Solve the inequality. Write the solution set in interval notation.
-$\frac { 9 h } { h - 1 } > - 7$
A) $\left( \frac { 7 } { 16 } , \infty \right)$
B) $\left( \frac { 7 } { 16 } , 1 \right)$
C) $\left( - \infty , - \frac { 7 } { 2 } \right) \cup ( 1 , \infty )$
D) $\left( - \infty , \frac { 7 } { 16 } \right) \cup ( 1 , \infty )$

Quizplus · Accepted Answer

To solve the inequality $\frac { 9 h } { h - 1 } > - 7$, first bring all terms to one side to get a common denominator:$$ \frac { 9 h } { h - 1 } + 7 > 0 $$$$ \frac { 9 h + 7(h - 1) } { h - 1 } > 0 $$$$ \frac { 9 h + 7h - 7 } { h - 1 } > 0 $$$$ \frac { 16h - 7 } { h - 1 } > 0 $$The critical points are found by setting the numerator and denominator equal to zero:$$ 16h - 7 = 0 \Rightarrow h = \frac{7}{16} $$$$ h - 1 = 0 \Rightarrow h = 1 $$These critical points divide the number line into three intervals: $(-\infty, \frac{7}{16})$, $(\frac{7}{16}, 1)$, and $(1, \infty)$. Test points from each interval in the inequality to determine where it is true.For $h < \frac{7}{16}$, choose $h = 0$:$$ \frac { 16(0) - 7 } { 0 - 1 } > 0 \Rightarrow \frac{-7}{-1} > 0 $$ which is true.For $\frac{7}{16} < h < 1$, choose $h = \frac{1}{2}$:$$ \frac { 16(\frac{1}{2}) - 7 } { \frac{1}{2} - 1 } > 0 \Rightarrow \frac{8 - 7}{-1/2} > 0 $$ which is false.For $h > 1$, choose $h = 2$:$$ \frac { 16(2) - 7 } { 2 - 1 } > 0 \Rightarrow \frac{32 - 7}{1} > 0 $$ which is true.Therefore, the solution set in interval notation is $(- \infty, \frac{7}{16}) \cup (1, \infty)$.

Solve the Inequality 9hh−1>−7\frac { 9 h } { h - 1 } > - 7h−19h​>−7

Solve the Inequality $\frac { 9 h } { h - 1 } > - 7$