How do you solve #(7x)/3 < 2#?

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Answer 1 · 2015-05-02T05:50:02

The answer is #x<6/7#.

Solve #(7x)/3<2# .

Multiply both sides by #3#.

#(7x)/cancel3 * cancel3/1 < 2 * 3# =

#7x<6# =

Divide both sides by #7#.

#(cancel 7x)/cancel7<6/7# =

#x<6/7#