Question #3b032

If you check again it appears that the question is.

#(5x)/6-3/10" and not " 5/x-3/10#

This accounts for the LCD of 30

Before we can add/subtract fractions they must have a #color(blue)"common denominator"# That is they have to be the same number on the denominator of both fractions.

We have to find a number then that is common to both 6 and 10, usually called the #color(blue)"lowest common multiple"#

One way to find it, is to divide the larger of the 2 numbers by the smaller and if it divides into it exactly , with a remainder of 0 then the larger number is the lowest common multiple. Repeat this process with multiples of the larger number until this happens.

#10÷6=1" remainder 4 so not 10 , now try 20"#

#20÷6=3" remainder 2 so not 20, now try 30"#

#30÷6=5" remainder 0"# thus 30 is the value.

The 2 fractions can now be expressed with a LCD of 30

#rArr(5x)/6xx5/5=(25x)/30" and " 3/10xx3/3=9/30#

Now that the fractions have the same denominator we can add/subtract the numerators but NOT the denominators.

#rArr(5x)/6-3/10=(25x)/30-9/30=(25x-9)/30#