Question

Solve the inequality \\(\\frac{2z+3}{z+2}\\leq 1\\)

Answer 1

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Solve the inequality \\(\\frac{2z+3}{z+2}\\leq 1\\)

Answer 2

SOLUTION

TO SOLVE

$\displaystyle \sf{ \frac{2z + 3}{z + 2} \leqslant 1}$

EVALUATION

Here the given inequality is

$\displaystyle \sf{ \frac{2z + 3}{z + 2} \leqslant 1}$

We solve it as below

$\displaystyle \sf{ \frac{2z + 3}{z + 2} \leqslant 1}$

$\displaystyle \sf{ \implies \: 2z + 3 \leqslant z + 2}$

$\displaystyle \sf{ \implies \: z \leqslant - 1}$

FINAL ANSWER

Hence the required solution is

$\displaystyle \sf{ z \leqslant - 1}$

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