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