Solve the inequality. Write the solution set in interval notation. $$\frac { 7 z + 1 } { - 3 } \geq z - 5$$
A) $\left[ \frac { 14 } { 5 } , \infty \right)$
B) $\left( - \infty , \frac { 7 } { 5 } \right)$
C) $\left[ \frac { 7 } { 5 } , \infty \right)$
D) $\left( - \infty , \frac { 7 } { 5 } \right]$

Question

Quizplus · Accepted Answer

To solve the inequality, multiply both sides by -3 (which reverses the inequality sign): $7z + 1 \leq -3(z - 5)$. Simplify to get $7z + 1 \leq -3z + 15$. Combine like terms: $10z \leq 14$. Divide by 10: $z \leq \frac{7}{5}$. The solution in interval notation is $(- \infty, \frac{7}{5}]$.

Solve the Inequality 7z+1−3≥z−5\frac { 7 z + 1 } { - 3 } \geq z - 5−37z+1​≥z−5

Solve the Inequality $\frac { 7 z + 1 } { - 3 } \geq z - 5$