How do you solve #5.6z+1.5<2.5z-4.7#?

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Answer 1 · 2017-04-02T11:36:33

#z<-2#

The inequality is still correct if we 'get rid' of the decimal by multiplying both sides by 10 giving:

#56z+15<25z-47#

Subtract #color(red)(25z)# from both sides

#color(green)(56z color(red)(-25z)+15" "<" "25zcolor(red)(-25z)-47)#

#" "color(green)(31z+15" "<" "-47#
..............................................................................................

Subtract #color(red)(15)# from both sides

#color(green)(31z+15color(red)(-15)" "<" "-47color(red)(-15))#

#" "color(green)(31z" "<" "-62#
................................................................................................

Divide both sides by #color(red)(31)#

#" "color(green)(31/(color(red)(31)) z" "<" "-62/(color(red)(31))#

But #31/31=1 and 1xx z = z#

#" "color(green)(z<-2)#