## please could someone help me get and understand this maths question?

(77 Posts)thanks, cant work it out at all

solve.

3a=3a-6 and a

... ....... ...

4 2 3

(the ... is meant to be a line, the "and" is a plus sign)

thanks

sorry, means

3a over 4 = 3a minus 6 over 2 + a over 3

a = 3, you need to multiply everything by 12 to get rid of the fractions and then rearrange it to work it out.

I'm not understanding where the "4 2 3" at the bottom fits in - is that meant to be part of an equation? The top bit works on its own

3a = 3a - 6 + a

subtract 3a from each side

0 = -6 + a

add 6 to each side

6 = 6 - 6 + a

get rid of the + and - 6 on the right

6 = a

so a=6

Make any sense?

ok mine makes no sense now

What MrsWobble says - when you have lots of fractions with different things on the bottom, you have to find something to multiply all the pieces by to make the fractions go away, then work from there.

So since you have things over 2,3,and 4, multiplying by 12 will make it much simpler to solve

thanks

i know i'm being annoying,but could someone possibly write it out for me the way you would in an exam?

sorry still am stuck

when i multiplied it all out by 12 i got 9a over 12 = 18a over 12???

multiply all by 12 and you get this:

36a = (36a-72)+12a

which can be simplified to 36a = 48a -72

which means that 72 = 12a

which means that a = 6 (72 divided by 12)

3a over 4 = 3a minus 6 over 2 + a over 3

3a/4 = (3a-6)/2+ a/3

I know that's not the answer, it's just a neater way to write the question.

and I suddenly realised that I'm too tired in the brain to work it out

3a/4 = 3a - 6/2 + a/3

multiply each bit by 12 -

3ax12/4 = 3ax12 - 6x12/2 + ax12/3

divide out the numbers where you can -

9a = 36a - 36 + 4a

sort out the a's on each side -

9a = 40a - 36

take away 9a from each side -

0 = 31a - 36

take away 36 from each side -

36 = 31a

divide both sides by 31 -

36/31 = a

you'll have to turn the /s into proper lines but you should be able to see where they belong?

Having done all that, I'm not sure if I've got the original question right - it seems odd having 6 over 2 as part of any equation, as you wouldn't usually put it like that. Can you see my working anyway?

I reckon nickelbabe's translation of the original question is probably right, though it doesn't give a nice answer which can be suspicious with some 'sanitised' maths teaching.

Multiplying through by 12 gives 9a = 18a - 36 + 4a

So 13a = 36

a = 36/13

I REALLY don't mean any offence by this, but wellwisher's answer may confuse you because she hasn't seen the denominators when multiplying through by 12.

I think mine is wrong, actually. Ignore

Finally - Schobe you have the same as me.

Hurrah for us - clearly we are right unless we've misunderstood the question!

I kept redoing it thinking it should have a nice simple answer but since mine matches yours we must both be right.

What age level is this?

What others have said, but with the working bit by bit -

3a/4 = (3a - 6)/2 + a/3

multiply each bit by 12 -

3ax12/4 = 12(3a - 6)/2 + ax12/3

divide out the numbers where you can -

9a = 6(3a - 6) + 4a

get rid of the brackets -

9a = 18a - 36 + 4a

sort out the a's on each side -

9a = 22a - 36

take away 9a from each side -

0 = 13a - 36

add 36 to each side -

36 = 13a

divide both sides by 13 -

36/13 = a

so 3a-6+a = 3a

so 3a-6=3a-a = 2a

so 3a=2a+6

so 3a-2a=6

so a=6

so 3x6 -6 +6 = 18-6+6 = 18 = 3 x 6

i get the same as amum

didn't see the fraction

lost the will to live

I think we've settled on -

3a/4 = (3a - 6)/2 + a/3

as the question - that would fit with the length of the .... lines on the original too, which I hadn't taken into account first time round

a = 36/13 i mean

sorry am more confused now

is a=6 or is it a=3

its age 13 homework BTW

a is not 6 or 3.

If the orignal question is as written by AMumInScotland her solution explains it clearly.

a is 36/13

Join the discussion

Registering is free, easy, and means you can join in the discussion, get discounts, win prizes and lots more.

Register nowAlready registered with Mumsnet? Log in to leave your comment or alternatively, sign in with Facebook or Google.

Please login first.