How do you solve #3x+1=1/2#?

2 Answers
Oct 8, 2015

#x=-1/6#

Explanation:

Subtract #1# from both sides

#3x+1-1=1/2-1#

#3x=-1/2#

And then divide both sides by #3#

#(3x)/3=(-1/2)/3#

#-1/2# divided by #3# can also be written as #-1/2# multiplied by #1/3#

#x=(-1/2)*1/3#

#x=-1/6#

OR in simpler terms* *

Bring #1# to the other side, making sure to change its sign

#3x=1/2-1#

#3x=-1/2#

And divide both numerators by #3#

#(3x)/3=-1/(2*3)#

#x=-1/6#

Oct 8, 2015

#x = -1/6#

Explanation:

People use the term "balance the equation" to describe how to solve this. Consider 5 = 5. Which is true. Write it as (5) = (5), the brackets are only there to show what is the original 'elements' of the equation as a group. Now consider the false statement that (5) -1 = 5. To balance this we need to write (5) -1 = (5) - 1. We have applied the same process to both sides.

Solving the original equation using the same idea:
The target is to get #x# on one side of the equals sign and everything else on the other.

First isolate all the elements with #x# in them. That is "#3x#"

#(3x+1)=(1/2)#

Step1. Subtract 1 from both sides giving:
#(3x+1) -1=(1/2)-1#
#3x=-1/2#

Step2. Now we need to separate the 3 from the #x# and move it to the other side of the equals sign. This is done by changing 3 into #1# as #1 times x = x#. Dividing by three is the same as multiply by #1/3#.

Divide both sides by three

#(3x) times 1/3 = (-1/2) times 1/3#

#3/3 times x = (-1 times 1)/(2 times 3)#

thus #x = -1/6#