How do you solve #-4(x-8)+(-x-3)=4#?

2 Answers
Anthony R.
Mar 24, 2017

#x=5#

Explanation:

Lets simplify this equation first by distributing the #-4# to #(x-8)#

#-4(x-8)->-4x+32#

Now lets combine like terms:
#color(blue)(-4x)+color(red)32color(blue)(-x)color(red)(-3)=color(red)4 ->color(blue)(-5x)+color(red)29=color(red)4#

Now we'll isolate the #x# by first subtracting 29 from both sides:

#color(blue)(-5x)+color(red)cancel(29-29)=color(red)(4-29)#

#color(blue)(-5x)=color(red)(-25)#

Then divide #5# on both sides:
#color(blue)cancel(-5/5x)=color(red)(-25/5)#

#color(blue)x=color(red)5#

We can verify the solution but substituting #5# for #x# in the original equation.

#-4((color(red)5)-8)-(color(red)5)-3=4#

#-4(-3)-(color(red)5)-3=4#

#12-(color(red)5)-3=4#

#7-3=4#

#4=4#

Meave60
Mar 24, 2017

#x=5#

Explanation:

Solve:

#-4(x-8)+(-x-3)=4#

Expand.

#-4x+32+(-x-3)=4#

Remove the parentheses.

#-4x+32-x-3=4#

Gather like terms.

#-4x-x+32-3=4#

Simplify.

#-5x+29=4#

Subtract #29# from both sides.

#-5x+cancel(29)-cancel(29)=4-29#

#-5x=-25#

Divide both sides by #-5#.

#(cancel(-5)x)/cancel(-5)=cancel(-25)^5/cancel(-5)^1#

#x=5#

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