Please help me with my easy Math SL homework?

[Gonzalo]

As ridiculous as this sounds, I have fraction related IB work that I just... cannot do.

I have always been bad at math and would love someone to help my with my fraction related math homework...

Combine

2/a² - 3/ab + 4/b²

1/x+5 + 2/3

1/x+5 + 2/x

1/x+5 + 2/x-3

Solve for x

4/x = x-6/x-4

4/x = 3x/x-3

x = x/3+5

3/x = x/3+5

3/x+2 = x/3 + 5/x+2

x+3/x-3 + x-3/x+3 = 18-6x/x²-9

PLEASE HELP ME.

I really wish I was made for Maths.

[Sonneteer_Trombonist]

I think the first part has to do with combining like terms. So what might be easiest is to convert all fractions to negative exponents. So 2/x becomes 2x^-1 and so forth. Then, see if any terms have the same variables behind it, and add the coefficients.

In the second part, you have a lot of denominators. So get rid of the denominators. You have to multiply each term by a common denominator. What you are going to do is look at every single denominator of a particular question, and then multiply each term by the denominators it doesn't have. In essence, this is like multiplying it by all the denominators, and then putting all the denominators on the bottom, and then cancelling with what it already had (does that sort of make sense? I'll do an example at the end).

Say for the first question, you have 4/x = (x-6)/(x-4) (sorry if this is the wrong equation, I am assuming brackets here. if the brackets are wrong, at least the steps should sort of be right).

So then you want the denominator to be x(x-4) for all terms so you can cancel them out. Well 4/x already has an x on the bottom but it doesn't have (x-4) so then it becomes 4(x-4)/x(x-4) and then (x-6)/(x-4) becomes x(x-6)/x(x-4). So then both the terms have x(x-4) in the denominator so you can cancel. Then you have 4(x-4)=x(x-6). Then you can expand and move all the x terms to one side.

Feel free to PM me if you have any questions, I'm not the greatest at putting math into words.

[Gonzalo]

THANK YOU SO MUCH.

It all makes a lot more sense.

(: