3y(y+4) = 3yy + 3y × 4 = 3y2+12y PLEASE can someone talk me thru how to do this sum?

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even though the answer is there I can't figure out how they got the answer, please can someone talk me thru it?

thanks

I'm not sure what is the question and what is the answer above?

Unless the question is 'does algebra bore you rigid', I don't know the answer.

What littlelapin said, you multiply the numbers in the () by 3y, so 3y x y= 3 y squared and 3y x 4= 12y. So you get 3y2 (sq.) + 12 y

its the middle bit that confuses me

3y(y+4) = 3yy + 3y × 4

sorry to be dim can someone talk me thru this?

the rules for multiplying this out are that you multiply the outside, the 3y by the first bit, the y to give 3y x y. Then you multiply the 3y by the last in the brackets to give 3y x 4, to give

I've put the brackets in to show you follow the bodmas rules of multiplying before adding.

3y(y+4) = (3y x y) + (3y x 4)

=3y2 + 12y

The whole thing is written strangely-it shows the progression- you wouldn't really write it like that, I don't think, at least we didn't in school.

The middle bit just shows how 3y is multiplied by y (hence 3yy) and by 4 (hence 3y x 4). In the last bit it's shown finished ie 3 y sq. + 12 y.

Basically you ultiply 3y by what is inside the (), first by 3y, then by 4.

3y(y + 4)

means 3y times ( whatever y plus 4 is )

and that is 3y times y plus 3y times 4

which is 3yy plus (3y x 4)

You can do it with real numbers if you want to see why it works

3 ( 1 + 4 )
= 3 x 5 = 15

or 3 x 1 plus 3 x 4
= 3 + 12 = 15

Does that help any?

It's written incorrectly:

When opening up the brackets:

3y(y+4)
can be written as 3yy + (3y X 4)

I think they have missed out rather essential brackets. Don't worry, it is their error and not yours.

Should mention that it is also a rather unnecessary step and makes it needlessly complicated.

The second brackets in
3yy + (3y x 4)
would make it clearer but are not essential, as multiplication takes priority over addition.

But they haven't!

3yy + (3y x 4)

is the same as

3yy + 3y x 4

As PlanetEarth said. In maths all division and multiplication in the equation is done first, before any addition and subtraction.

This is known as the distributive law of multiplication.

But, of course, giving it a name is no help in understanding what it means.

The limitations of how you can type stuff into a message on MN restrict the way you can type maths stuff, and possibly even add to confusion and misunderstandings.

This gives a good visual idea as to what's going on.

www.mathsteacher.com.au/year7/ch05_algebra/09_dist/law.htm

thanks to everyone who replied, TBH I'm still confused but will have a look at the above link

The basic principle is that 6x + 4x = 10x. So:

(6 + 4) x

is the same as

10x

To make that general, replace the 6 and 4 with a and b and we get:

(a + b) x = ax + bx

That is why

3y (y + 4) = 3yy + 12y

When you have a number outside the brackets it means that number times everything in the brackets.

But you can't times everything in the brackets all at once because the things in the brackets are separated by a plus sign, so you have to times each of them individually by the bit outside the brackets.

So...

First you do the 3y from outside the brackets times the first bit from inside the brackets, which is just a y.

When you times 3y by y, you get 3ysquared. In the way your equation has been set out, this is described in the middle stage as 3yy, because yy means ysquared or y times y. I think this may be why you are confused, because when you move on to the final stage this 3yy has not changed, but they have changed the way it is writen to 3y2.

So at this stage you have done the first half of the sum but not the second half. In your equation, this middle stage has been written out as 3yy + 3y x 4

They have written it like that to show that you have already got your 3yy but now you have to do the other half of the sum which will be 3y times 4.

After doing that first multiplication, you next need to times the 3y outside the brackets with the other part inside the brackets, which is 4. So you do 3y times 4. 4 lots of 3 y are 12y. So the answer to that bit is 12 y.

So now you have the answers to the two times sums - the first bit is 3ysquared or 3yy and the second bit is 12y.

So your final thing to do is just to add those together, giving you the final answer of 3ysquared plus 12y.

I really hope this helps - my Sam really struggled so I mastered it while he was doing algebra!!

3y aye tiddly aye tye about covers it