How do you evaluate #1/3+1/2+2/3#?

4 Answers
Jul 15, 2018

Answer:

#color(maroon)(=> 9/6 = 1 1/2#

Explanation:

#1/3 + 1/2 + 2/3#

As there is no common factor between 2 & 3, 2*3 = 6 is the L C M.

#=> (1/3) * (2/2) + (1/2)* (3/3) + (2/3)*(2/2)# as 6 is the L C M of 2 & 3.

#=> (2/6) + (3/6) + (4/6)#

#=> (2 + 3 + 4) / 6#

#=> 9/6#

#=> 1 (cancel(3)^color(red)(1) / cancel(6)^color(red)(2))= 1 1/2#

Jul 15, 2018

Answer:

See a solution process below:

Explanation:

To add fractions they must be over a common denominator.

To change the denominator without changing the value of the fraction we can multiply each fraction by a form of #1#

#1/3 = 2/2 xx 1/3 = (2 xx 1)/(2 xx 3) = 2/6#

#1/2 = 3/3 xx 1/2 = (3 xx 1)/(3 xx 2) = 3/6#

#2/3 = 2/2 xx 2/3 = (2 xx 2)/(2 xx 3) = 4/6#

We can now rewrite the problem and add the fractions as:

#2/6 + 3/6 + 4/6 = (2 + 3 + 4)/6 = 9/6#

We can reduce the fraction as:

#9/6 = (3 xx 3)/(3 xx 2) = (color(red)(cancel(color(black)(3))) xx 3)/(color(red)(cancel(color(black)(3))) xx 2) = 3/2#

If necessary, we can convert this improper fraction into a mixed number:

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

Jul 15, 2018

Answer:

#3/2#

Explanation:

Recall that to add fractions, we must have like denominators. We can start off by adding the fractions with the denominator of #3# to get

#3/3+1/2=1 1/2#, or #1.5#.

A more systematic way would be to get a common denominator of #6#, since this is the LCD of the fractions.

To get a denominator of #6#, we can multiply the first by #2/2#, the second by #3/3#, and the third by #2/2#. We now have

#2/6+3/6+4/6#

Adding the numerators, we get

#9/6#, or #3/2#, or #1 1/2#, or #1.5#.

Hope this helps!

Answer:

#3/2#

Explanation:

#1/3+1/2+2/3#

#=\frac{1\cdot 2+1\cdot 3+2\cdot 2}{6}#

#=\frac{2+3+4}{6}#

#=\frac{9}{6}#

#=\frac{3\cdot 3}{3\cdot 2}#

#=3/2#