The Commutative property: Using fractions

$4\frac{2}{3}+2\frac{2}{7}+3\frac{1}{3}+1\frac{3}{7}=\text{?}$

Given that this is an exercise with only addition operation, we can change the order of the numbers.

We organize the exercise in a way that we can obtain a pair that gives us an integer.

Keep in mind that there is a pair of fractions that if we add them we will obtain an integer:

$4\frac{2}{3}+3\frac{1}{3}+2\frac{2}{7}+1\frac{3}{7}=$

We solve the exercise from left to right:

$4\frac{2}{3}+3\frac{1}{3}=$

$4+3=7$

$\frac{2}{3}+\frac{1}{3}=\frac{3}{3}=1$

$7+1=8$

Now we obtain the exercise:

$8+2\frac{2}{7}+1\frac{3}{7}=$

We leave the 8 aside and add the rest of the exercise:

$2\frac{2}{7}+1\frac{3}{7}=$

$2+1=3$

$\frac{2}{7}+\frac{3}{7}=\frac{5}{7}$

Now we obtain the exercise:

$8+3+\frac{5}{7}=11\frac{5}{7}$

$11\frac{5}{7}$

^{$5.2-7.4+12.4+3.2+6.6=\text{?}$}

To make things simpler, we use the substitution property and arrange the exercise in the following manner:

$5.2+12.4-7.4+3.2+6.6=$

Keep in mind that the subtraction operation here gives us a whole number:

$12.4-7.4=5$

Now we obtain the exercise

$5.2+5+3.2+6.6=$

We solve the addition exercise:

$3.2+6.6=9.8$

And we obtain the exercise:

$5.2+5+9.8=$

We arrange the exercise using the substitution property to make fiding the solution simpler:

$5.2+9.8+5=$

We solve the exercise from left to right:

$5.2+9.8=15$

$15+5=20$

20

$4\frac{1}{4}\cdot3\frac{4}{9}\cdot3\frac{1}{17}=\text{?}$

Multiply each fraction as follows:

Multiply the whole number by the denominator of the fraction and add the number in the numerator of the fraction.

That is:

$4\frac{1}{4}=\frac{4\times4+1}{4}=\frac{16+1}{4}=\frac{17}{4}$

$3\frac{4}{9}=\frac{9\times3+4}{9}=\frac{27+4}{9}=\frac{31}{9}$

$3\frac{1}{17}=\frac{17\times3+1}{17}=\frac{51+1}{17}=\frac{52}{17}$

Now we get the exercise:

$\frac{17}{4}\times\frac{31}{9}\times\frac{52}{17}=$

We simplify the 17 and get:

$\frac{31\times52}{4\times9}=\frac{52}{4}\times\frac{31}{9}=13\times\frac{31}{9}=\frac{403}{9}=44\frac{7}{9}$

$44\frac{7}{9}$

Damian is training for a race.

The first day he ran 3.4 km..

The second day he ran round trip for 1.18 km..

On the third day Damian ran 2.6 km...

How many kilometers in total did Damian run during the three days of training?

Let's consider the number of kilometers Damian ran each day

On the second day it is written "round trip", that is, twice.

Therefore, we write the following exercise:

$3.4+2\times1.18+2.6=$

We solve the multiplication exercise:

$3.4+2.36+2.6=$

We arrange the exercise using the commutative property, to make it more convenient to solve:

$3.4+2.6+2.36=$

We solve the exercise from left to right:

$3.4+2.6=6$

$6+2.36=8.36$

8.36

$5.25\cdot\frac{2}{3}\cdot\frac{7}{21}=\text{?}$

We rewrite the decimal fraction in the form of a mixed fraction:

$5.25=5\frac{1}{4}$

Now we get the exercise:

$5\frac{1}{4}\cdot\frac{2}{3}\cdot\frac{7}{21}=$

We write the mixed fraction as a simple fraction:

$5\frac{1}{4}=\frac{4\times5+1}{4}=\frac{20+1}{4}=\frac{21}{4}$

Now we obtain:

$\frac{21}{4}\cdot\frac{2}{3}\cdot\frac{7}{21}=$

We simplify the 21 and obtain:

$\frac{7}{4}\cdot\frac{2}{3}=\frac{14}{12}=\frac{12+2}{12}=1\frac{2}{12}=1\frac{1}{6}$

$1\frac{1}{6}$

Complete the exercise:

$0.8+\frac{2}{10}-\frac{3}{2}\times\frac{4}{2}+\frac{1}{2}=$

1.5-

Complete the the exercise:

$0.2+\frac{2}{4}-\frac{1}{8}\times2+0.4=$

$0.85$

Complete the exercise:

$\frac{1}{5}+0.4-\frac{2}{9}\colon\frac{1}{3}+1=$

$1.6-\frac{2}{3}$

Complete the exercise:

$\frac{2}{4}+\frac{2}{5}\times\frac{5}{4}-0.2+0.4=$

1.2

$(\frac{1}{4}+\frac{7}{4}-\frac{5}{4}-\frac{1}{4})\cdot10:7:5=\text{?}$

$\frac{1}{7}$

$\frac{1}{4}\times(\frac{1}{3}+\frac{1}{2})=$

$\frac{5}{24}$