How do you solve #10+ x < 4x + 1< 33#?

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Answer 1 · 2018-05-02T01:12:20

#x>3# and #x<8#

So when we have inequalities as such, we split it.
#10+x < 4x+1 < 33# becomes

#10+x< 4x+1# and #4x+1 < 33#

Then we solve the inequality as normal

#10+x< 4x+1#

#10-1 < 4x-x#

#9 < 3x#

therefore, #9/3 < x# and #3 < x#.

or #x > 3# (x is greater than 3)

then:

#4x+1 < 33#

#4x < 33-1#

#x < 32/4#

#x < 8# (x is less than 8))

hence, #x> 3# and #x <8#