How do you solve # (3/5)x ≤ 10 + (1/5)x#?

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Answer 1 · 2015-05-05T04:27:35

The answer is #x<=25# .

Solve #3/5x<=10+1/5x# as if there was an equal sign between the two sides instead of an inequality.

Subtract #1/5x# from both sides to get #x# on one side.

#2/5x<=10#

Multiply both sides by #5/2# to cancel #2/5#. (This is the same as dividing by #2/5# .)

#cancel 2/cancel 5x*cancel5/cancel 2<=10*5/2# =

#x<=50/2# =

#x<=25#