How do you solve #-6.4x = 128.0#?

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Answer 1 · 2017-03-03T08:05:27

#x=20#

#-6.4x=128.0#

Multiply both sides by #10# to change the decimal into a whole number.

#-64x=1280#

Divide both sides by #-64#.

#(-64x)/-64=1280/-64#

The minus signs on the left cancel each other, so:

#(64x)/64=-1280/64#

#(cancel64x)/(cancel64)=1280/64#

#x=1280/64#

Since #64# goes into #1280# #20# times:

#x=20#