#1

If you're doing (b+5ax)/x, would your answer be b+5a, or am I skipping something here?

This little fundamental problem is setting me back a bit, and I can't seem to find any answers on the internet.

This little fundamental problem is setting me back a bit, and I can't seem to find any answers on the internet.

#2

b/x+5a is it simplified i believe

#3

If you're doing (b+5ax)/x, would your answer be b+5a, or am I skipping something here?

This little fundamental problem is setting me back a bit, and I can't seem to find any answers on the internet.

i don't know the answer, but i do know that's wrong.

#4

b/x+5a

you divide both terms by x.. the b becomes a fraction. in the 5ax, the x's cancel out

you divide both terms by x.. the b becomes a fraction. in the 5ax, the x's cancel out

#5

b/x+5a

you divide both terms by x.. the b becomes a fraction. in the 5ax, the x's cancel out

Okay, thanks. That was actually the other answer I was thinking it would be...

Anyone dare to tell this man he is wrong? I'd really like a few opinions.

#6

yep edawgs right

#7

hes dead on

#8

hes right because if you divide both terms by x then its initially b/x + 5ax/x and that then simplifies to b/x + 5a

#9

man i hate math.

im pretty sure hes right though

im pretty sure hes right though

#10

Nirvana guy beat me to it, but yea

its b/x+5a.

i guess the next qustion is,

whos in algerbra 1?

its b/x+5a.

i guess the next qustion is,

whos in algerbra 1?

#11

Well, if it gives you any peace of mind just know that you will rarely need to do that kind of stuff when you get into higher math courses because b/x+5a is just as hard to work with as (b+5ax)/x. But I guess you still need to know how to do it.

#12

Okay, thanks. That was actually the other answer I was thinking it would be...

Anyone dare to tell this man he is wrong? I'd really like a few opinions.

You don't need any, he's right.

#13

YES

The answer is b/5 + 5a

The answer is b/5 + 5a

#14

I ended up with the area between the paraboloid x^2 + z^2 = 1 and the sphere 3(x^2+y^2+z^2) = 27. I must be doing something wrong.