The following inequality has a solution in the form x≤A.\displaystyle {x}\le{A}. Solve the inequality and place the correct value of A\displaystyle {A} into the box.

19x+12≤9+12x\displaystyle {19}{x}+{12}\le{9}+{12}{x}

Answer: x≤\displaystyle {x}\le
Enter your answer as an integer or a reduced fraction in the form A/B.
 

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