How do you solve #\frac { 1} { 3} [ 2( x - 8) + 1] = 19#?

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Answer 1 · 2018-06-09T07:03:05

#x=36#

As per the question, we have

#1/3[2(x-8)+1]=19#

#:.1/3[2x-16+1]=19#

#:.1/3[2x-15]=19#

#:.1/3(2x-15)xx3=19xx3# ... [Multiplying #3# on both sides]

#:.1/cancel3(2x-15)xxcancel3=57#

#:.2x-15=57#

#:.2x-15+15=57+15# ... [Adding #15# on both sides]

#:.2x=72#

#:.x=36#

Hence, the answer.

Answer 2 · 2018-06-09T07:07:33

Here is an algebraic rewrite to make it simpler

First rewrite this in to simplest terms:
#1/3[2(x-8)+1]=19 -> 2(x-8)+1 = 19/3#,

#2(x-8)+1=19/3 ->2(x-8)= 19/3-1#,

#2(x-8)=19/3 -3 -> x-8=(19/3-1)/2#,

#x-8=(19/3-1)/2 -> x=(19/3-1)/2 +8#

Now you can evaluate the expression and solve for x.