gcse maths help

How might you make the LHS a single fraction?

If you multiply the denominators, what will happen to the numerators?

if i multiply (x-2)(x+3) would the LHS become 6(x+3) / x^2 + x - 6 and 2(x-2) / x^2 + x - 6 = 1,,,,, n i also multiply the 1 by the denominators too??? idk

So, when solving this type of a question, the first thing you could try, is to see if you can simplify the left hand side of the equation into a fraction with a single(common) denominator.

Have you studied/have worked out any calculations like this before in your class?

Let me know if need further help to proceed ahead

You need to multiply all terms and simplify - the LHS numerators look correct

so 6x + 18 / x^2 + x - 6 and 2x - 4 / x^2 + x - 6 = 1,,,, what do i do now lol

You haven't simplified the left or multiplied on the right

so the left would simplify to 4x - 1 / x^2 + x - 6 but on the right do i multiply the 1 by (x-2)(x-3) which would make the denominator the same as that on the left side??

I don't agree with your simplifying and you are making this far moe difficult than need be...
If you multiply through by the common denominator you get:

6(x + 3) - 2(x - 2) = (x + 3)(x - 2)

so the solutions r 7 n -4 if u work out using completing the square,,,, i got itttt thank u sm!!! <3333

i tried doing thatttt it's too hard,, i'm homeskled but i did try sum like this b4 i think? but yesssss i need help omg

Wh


Where exactly you find it hard? Let me know so that I could help you on that

so the solutions r 7 n -4 if u work out using completing the square,,,, i got itttt thank u sm!!! <3333