How do you solve #82= - 2( 5v - 3) - 8#?

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Answer 1 · 2017-10-09T01:18:11

#v=-8.4#

When solving equations with a variable, you want to isolate for that variable, which in this case, is #v#.

First, add #8# to both sides, which will give you:

#90=-2(5v-3)#

Then, distribute the #-2# on the RHS using the distributive property:

#90=-2(5v)-2(-3)#

Multiply the terms on the RHS:

#90=-10v+6#

Subtract #6# from both sides:

#84=-10v#

Divide both sides by #-10#:

#v=-8.4#

Answer 2 · 2017-10-09T01:18:48

#v=-(42/5)=-8.4#

#-2(5v-3)-8=82#
#-10v+6-8=82#
#-10v=82+2#
#v=-(84/10)=-(42/5)=-8.4#

Answer 3 · 2017-10-09T01:27:56

#v = -42/5 or 8.4#

#82 = -2(5v - 3) - 8#

Expressing the bracket

#82 = -10v + 6 - 8#

Simplifying

#82 = -10v - 2#

Collecting like terms

#82 + 2 = -10v#

#84 = -10v#

Divide both sides by #color(blue)(-10)#

#84/color(blue)(-10) = (-10v)/color(blue)(-10)#

#84/-10 = (cancel(-10)v)/cancel(-10)#

#-84/10 = v#

#v = cancel(-84)^42/cancel(10)_5#

#v = -42/5 or 8.4#