Question: i couldn't find in the book on how to solve

x+6/7 = 3/5

i know its simple but i don't remember how and the book never explains step by set on how to solve problems and i don't have a teacher to ask because i do work through mail so i cant ask questions vary easily
thx for the help

This is an algebraic problem - you need to "solve for x" which means finding out what number replaces x to make the equation right.

The first step is to try to isolate x on one side of the equal sign (it doesn't really matter which side). Since the left-hand-side (LHS) of the equal sign is "x + 6/7", you want to somehow get rid of the "+ 6/7". The easiest way is to just subtract it, because if you add something and then immediately subtract it, you get the original value:

x + 6/7 - 6/7

is just
x

One of the rules of working with an equation is that whatever you do to one side you must do to the other. So if we subtracted 6/7 from the LHS, we must at the same time subtract 6/7 from the RHS:

3/5 - 6/7

To solve this as a fraction, you have to use a common denominator. In this case, 35 works well. You get that by multiplying the bottom by 7 and just as with the sides of an equation, whatever you do to the bottom you have to do to the top.

So multiply the top and bottom of the first fraction (3/5) by 7 to make the denominator 35.

(3 x 7) / (5 x 7) = 21 / 35

and to change (6/7) to something over 35, you multiply top and bottom by 5:

(6 x 5) / (7 x 5) = 30 / 35.

Now these new fractions can replace the original 3/5 - 6/7 and you can subtract the numerators:

21/35 - 30/35 = (21 - 30) / 35 = -9 / 35

Now you have x on the LHS and -9/35 on the RHS.

Sorry to go so slowly through this Cavell, but I think other students who are trying hard to understand fractions and just starting algebra will also be helped by your problem.

Stephen La Rocque>