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How do you solve #5(-3x-2)-(x-3) = -4(4x+5)+13#?

How do you solve #5(-3x-2)-(x-3) = -4(4x+5)+13#?
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ReportAnswer #1

#x# has any value.

Explanation:

Remove the brackets first using the distributive law.

#5(-3x-2) -(x-3) = -4(4x+5)+13#

#-15x-10-x+3 = -16x-20+13#

#-16x -7 = -16x-7" "larr# both sides are identical.

#-16x+16x = -7+7" "larr# this leads to

#0=0" "larr# true, but there is no #x# to solve for.

This is an identity which means that it will be true for any value of #x#

answered by: EZ as pi
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How do you solve #5(-3x-2)-(x-3) = -4(4x+5)+13#?
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