Helps with Math?

[theelitegamer]

Hey what's sup guys, I know this is a tech forum and all but, I kindof need help. I wanna know how to solve one of the problems in the picture, if anyone knows how to do so, can you please help me solve one of them, so I know how to do that and be on my way?

[steffen_anywhere]

Not actually hard.

Just make the denominator the same by multiplying them.

Then multiply each numerator with the other fraction's denominator as well. Then add the numerators and youre done.

Answer is the third from the top

[Ciccioo]

link to colorful explaination

[theelitegamer]

link to colorful explaination

Thanks matey

[theelitegamer]

Not actually hard.

Just make the denominator the same by multiplying them.

Then multiply each numerator with the other fraction's denominator as well. Then add the numerators and youre done.

Answer is the third from the top

I know right? it sounds easy but when you get around to doing it, your like wtf is this?

[WiiManic]

Been answered, but screw it, was waiting for phone to sync.

Multiply by the denominator like said, answer is third one.

 

I apologise for handwriting....

[oc0110]

what steffen said for the first question

 

and the process for the second question:

 

(r+2)/(r+4) - 3/(r+1) = [(r+2)(r+1) - 3(r+4)]/[(r+4)(r+1) = (r^2 + 2r + r + 2 - 3r -12)/(r^2 + 4r + r + 4) = (r^2 - 10)/(r^2 + 5r + 4)

[helping]

I wish I could go back to doing this type of comfort math, would help my self esteem so much and make me feel good about myself

[oc0110]

what u have to do is make a common denominator so multiply the bottom denominators and make a new denominator. Then multiply the missing denominator for each fraction and subtract or add.

 

ex:

(x)/(x-2) - (5x^2 + 1)/(3)

--> make a common denominator

(x-2)*(3) = 3(x-2) = 3x - 6 <-- this is ur new denominator

 

now from the fraction:(x)/(x-2) it's missing the denominator 3 from the fraction so multiply both the numerator and the denominator

--> (3x)/(3x-6) <-- the 3x -6 is the common denominator

 

do the same to the other side

since x-2 is missing

multiply top and bottom with x-2

--> and u'll get [(5x^2 + 1)(x-2)] / 3(x-2)

--> = (5x^3 + x -10x^2 -2)/(3x-6)

 

now combine them:

(3x)/(3x-6) - (5x^3 + x -10x^2 -2)/(3x-6)

= (-5x^3 +10x^2 +2x +2)/(3x-6)

 

 

That should explain it pretty well.

ugh miss this kind of math after doing IB HL math...