The original steps are 1.To find the LCD(least common denominator) 2.Multiply both sides by the LCD 3.Simplify and 4.Solve. I have trouble multiplying and simplifying polynomials and trinomials. I tried one problem using cross multiplication and it worked but I don’t know if this is an absolute way to solve the equations. Please help if you understand this? >.< No rude comments please

Right, so you need to be able to solve, for example, x/12-1/3+1/2x=0. It makes things easier if we multiply up both sides of the equation to get rid of the fractions (although it’s not strictly necessary to do so). The least common denominator is just the smallest number (or algebraic expression, as in this case) we can use to achieve this. In any case, multiplying both sides by all of the product of the denominators will do fine as well. In this case, that would mean multiplying both sides by 36x (=12*3*x). However, 12x will be the LCD. If we use this, the equation becomes 12x(x/12-1/3+1/x)=0 and this simplifies to x^2-4x+12=0, which can be solved by factorisation.

The only problem with cross multiplication arises when we have a variable in the denominator and an inequality. This is because if we multiply by the variable (eg x, as in the equation mentioned) we don’t know whether we are multiplying by a positive or a negative number. As far as I know, we then have to consider both possibilities separately.

So if we have x/12-1/3+1/x<0, then, assuming x is positive, 12x(x/12-1/3+1/x)<0 and x^2-4x+12<0. After factorising the left hand side, we know that we must have one positive and one negative factor in order to get the required negative value and this is only possible when -2<x<6. But since this is based on the assumption that x is positive, we can only infer that 0<x<6 satisfies the inequality.

Assuming instead that x is negative gives us 12x(x/12-1/3+1/x)>0 (because the greater of the two sides will (in any case) have become the lesser after we multiply both by a negative number), so that x^2-4x+12>0. This leads to the factors having to have the same sign, which is only possible when x>6 or x<-2. But like before, since we assumed that x was negative (this time) we can only infer that x<-2 satisfies the inequality (ie, is true given that the inequality is true).

The final possibility to consider is that x is zero, but since this results in 1/0 within the left hand side of the original inequality, is cannot satisfy it. Hence, we can conclude that either 0<x<6 or x<-2.