How do you evaluate #(2x+5)/ 3 = (x-5)/4# when #x# is (a) -7, (b), -4, (c) -2 and (d) -1?

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Answer 1 · 2017-12-25T01:00:29

#x=-7#

Given -

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

At #x=-7#

#(2(-7)+5)/3=((-7)-5)/4#
#(-14+5)/3=(-7-5)/4#
#(-9)/3=(-12)/4#
#-3=-3#

At #x=-4#

#(2(-4)+5)/3=((-4)-5)/4#
#(-8+5)/3=(-4-5)/4#
#(-3)/3!=(-9)/4#

At #x=-2#

#(2(-2)+5)/3=((-2)-5)/4#
#(-4+5)/3=(-2-5)/4#
#1/3!=(-7)/4#

At #x=-1#

#(2(-1)+5)/3=((-1)-5)/4#
#(-2+5)/3=(-1-5)/4#
#(3)/3!=(-6)/4#