Solve the inequality. $\frac { z - 1 } { 7 } + \frac { z + 1 } { 6 } \frac { 1 } { 42 }$ A) $z \geq 0$ B) $z 0$ C) $z \leq 0$ D) $z

Question

Solve the inequality. $\frac { z - 1 } { 7 } + \frac { z + 1 } { 6 } > \frac { 1 } { 42 }$
A) $z \geq 0$
B) $z > 0$
C) $z \leq 0$
D) $z < 0$

Quizplus · Accepted Answer

To solve the inequality $\frac { z - 1 } { 7 } + \frac { z + 1 } { 6 } > \frac { 1 } { 42 }$, first find a common denominator for the fractions, which is 42. This gives us $\frac { 6(z - 1) } { 42 } + \frac { 7(z + 1) } { 42 } > \frac { 1 } { 42 }$. Simplifying, we get $\frac { 6z - 6 + 7z + 7 } { 42 } > \frac { 1 } { 42 }$, which simplifies further to $\frac { 13z + 1 } { 42 } > \frac { 1 } { 42 }$. Multiplying both sides by 42 to eliminate the denominator, we get $13z + 1 > 1$, which simplifies to $13z > 0$. Dividing both sides by 13, we find $z > 0$.

Solve the Inequality z−17+z+16>142\frac { z - 1 } { 7 } + \frac { z + 1 } { 6 } > \frac { 1 } { 42 }7z−1​+6z+1​>421​

Solve the Inequality $\frac { z - 1 } { 7 } + \frac { z + 1 } { 6 } > \frac { 1 } { 42 }$