Amount of Money Withdrawn

Algebra 2

Re: Amount of Money Withdrawn

Postby Guest » Mon May 02, 2016 8:14 pm

Your last 2 posts are not correct.

As I said in my last post..... You need to read the sum1.jpg .....

if you read it.....

..... You will see where I multiplied both sides by A......indicated by ..... X A gives......

Then there was a slight bit of rearranging and it was left for you to cross multiply and finish the remaining steps....
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Re: Amount of Money Withdrawn

Postby Guest » Mon May 02, 2016 9:15 pm

6A / (5 - A) = (6 + A) / 1

6A * 1 = (5 - A) * (6 - A)

6A = 30 + 5A - 6A - A^2

I can't solve.
Guest
 

Re: Amount of Money Withdrawn

Postby Guest » Tue May 03, 2016 5:20 am

6A / (5 - A) = (6 + A) / 1

6A * 1 = (5 - A ) * (6 + A) ........OK

You have got it correct up to here in the post before your last one. This is the cross multiplying bit.

You need to be careful with the signs, especially when multiplying the brackets containing 2 terms.

You will end up with a quadratic equation to solve.

Like all quadratics, you make it equal to zero and solve by factorising, OR completing the square, OR using the formula.

You will find this quadratic is very simple and can be solved by factorising very easily.
Guest
 

Re: Amount of Money Withdrawn

Postby Guest » Tue May 03, 2016 9:35 am

I can't solve it.
Guest
 

Re: Amount of Money Withdrawn

Postby Guest » Tue May 03, 2016 8:02 pm

the RHS is just 2 brackets that need multiplied......you had this done on a previous post but got signs wrong in places.

the LHS is just 6A x 1 which is just 6A
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Re: Amount of Money Withdrawn

Postby Guest » Tue May 03, 2016 10:09 pm

I still can't.
Guest
 

Re: Amount of Money Withdrawn

Postby Guest » Wed May 04, 2016 5:10 am

Do you read the previous post that you do...???
Below is a previous post you posted.....
What you have done here is OK except for the last line, you did not deal with the signs correctly
You should associate the sign with the letter or number following it to the right
eg. -A .....this is minus A ....... +A this is plus A
- 6 ..... this is minus 6 ..... +6 this is plus 6.
a minus number times a minus number gives a plus result ..... -4 x -2 = -8 and -F x -G = -FG

6A / (5 - A) = (6 + A) / 1

6A * 1 = (5 - A) * (6 - A) .....this is cross multiplying...OK

6A = 30 + 5A - 6A - A^2 ....you have got the signs wrong here, the multiplying method is OK otherwise.

You just need to do the last line again, then arrange in standard form of quadratic equation
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Re: Amount of Money Withdrawn

Postby Guest » Wed May 04, 2016 10:13 am

" Do you read the previous post that you do...??? " Yes.

I don't know how.
Guest
 

Re: Amount of Money Withdrawn

Postby Guest » Wed May 04, 2016 3:02 pm

...............................
Please note I posted this line wrong in earlier post........ it should have been +8 and +FG...sorry about that...

a minus number times a minus number gives a plus result ..... -4 x -2 = +8 and -F x -G = +FG
...........................................................

Do this bit below again (shown bold) and correct the signs

6A / (5 - A) = (6 + A) / 1

6A * 1 = (5 - A) * (6 - A) .....this is cross multiplying...OK

6A = 30 + 5A - 6A - A^2 ....you have got the signs wrong here, the multiplying method is OK otherwise.

You just need to do the last line again, then arrange in standard form of quadratic equation
Guest
 

Re: Amount of Money Withdrawn

Postby Guest » Wed May 04, 2016 3:21 pm

6A = 30 + 5A - 6A - A^2

Redoing: 6A = 30 + 5A - 6A - A^2

6A = 30 - A - A^2


I can't do it.
Guest
 

Re: Amount of Money Withdrawn

Postby Guest » Wed May 04, 2016 6:17 pm

It looks as though you just simply wrote what you had already done down again on your last post...

This is what I had asked you to do on an earlier post........
.............................................
6A / (5 - A) = (6 + A) / 1

6A * 1 = (5 - A) * (6 - A) .....this is cross multiplying...OK

6A = 30 + 5A - 6A - A^2 ....you have got the signs wrong here, the multiplying method is OK otherwise.

.........................................
It looked like you had used the correct method of multiplying, but got the signs wrong.......
Here are the steps you would have taken......

6A * 1 = 6A ......you got this correct.
5 * 6 = 30 ......you got this correct.
5 * -A = ? .....this is a plus x a minus and that gives minus ......unlike signs gives minus

-A * 6 = ? ..... minus * plus gives minus
-A * -A = ? ...... minus * minus gives ??? .....like signs give minus

You should be able to tidy this up now and put in the form of a quadratic equation
Guest
 

Re: Amount of Money Withdrawn

Postby Guest » Wed May 04, 2016 6:38 pm

6A * 1 = 6A
5 * 6 = 30
5 * -A = -5A

-A * 6 = -6A
-A * -A = -A^2


6A = 30 - 5A - 6A - A^2 Just what I have before following your comments. I don't get it.
Guest
 

Re: Amount of Money Withdrawn

Postby Guest » Wed May 04, 2016 7:53 pm

Sorry, I had an error in my last post...............
...........................
It looks as though you just simply wrote what you had already done down again on your last post...

This is what I had asked you to do on an earlier post........
.............................................
6A / (5 - A) = (6 + A) / 1

6A * 1 = (5 - A) * (6 - A) .....this is cross multiplying...OK

6A = 30 + 5A - 6A - A^2 ....you have got the signs wrong here, the multiplying method is OK otherwise.
.........................................
It looked like you had used the correct method of multiplying, but got the signs wrong.......
Here are the steps you would have taken......

6A * 1 = 6A ......you got this correct.
5 * 6 = 30 ......you got this correct.
5 * -A = ? .....this is a plus x a minus and that gives minus ......unlike signs gives minus

-A * 6 = ? ..... minus * plus gives minus
-A * -A = ? ...... minus * minus gives ??? .....like signs give Plus ......Sorry typo error, I was wrong here...

You should be able to tidy this up now and put in the form of a quadratic equation
..................................................

6A * 1 = 6A
5 * 6 = 30
5 * -A = -5A

-A * 6 = -6A
-A * -A = +A^2 ........This should be plus, sorry abut typo error mentioned above


6A = 30 - 5A - 6A + A^2 Just what I have before following your comments. I don't get it. Yes this is it now OK

6A = 30 + 5A - 6A - A^2 ....you have got the signs wrong here, the multiplying method is OK otherwise. This was your earlier post
Guest
 

Re: Amount of Money Withdrawn

Postby Guest » Wed May 04, 2016 8:03 pm

Now what do I do ?
Guest
 

Re: Amount of Money Withdrawn

Postby Guest » Thu May 05, 2016 6:18 pm

Please reply.
Guest
 

Re: Amount of Money Withdrawn

Postby Guest » Thu May 05, 2016 7:55 pm

You need to go back to the jpeg I posted an make sure you are doing the correct sum.........

There are numerous posts since that that have wrong signs in the sum and also in the multiplying so it is confusing things to follow from these posts.

In some cases I followed on from your working but there was actually a wrong sign in the sum you were starting with...????

This is a copy of a previous post that is correct up to the cross multiplying bit. You need to multiply out the brackets and get the signs correct, then rearrange it in quadratic form and solve as explained in post below......

...............................................copy of earlier post below...
6A / (5 - A) = (6 + A) / 1

6A * 1 = (5 - A ) * (6 + A) ........OK

You have got it correct up to here in the post before your last one. This is the cross multiplying bit.

You need to be careful with the signs, especially when multiplying the brackets containing 2 terms.

You will end up with a quadratic equation to solve.

Like all quadratics, you make it equal to zero and solve by factorising, OR completing the square, OR using the formula.

You will find this quadratic is very simple and can be solved by factorising very easily.

......................................................
Guest
 

Re: Amount of Money Withdrawn

Postby Guest » Thu May 05, 2016 8:38 pm

6A * 1 = (5 - A ) * (6 + A)

6A = 30 + 5A - 6A - A^2

I don't know how to solve.
Guest
 

Re: Amount of Money Withdrawn

Postby Guest » Fri May 06, 2016 7:05 pm

Just never mind. I have wasted enough of your time. Thanks anyway.
Guest
 

Re: Amount of Money Withdrawn

Postby Guest » Fri May 06, 2016 7:11 pm

If you are being asked to do questions like this then you must have some knowledge of quadratics.
This is a worded question where you have to derive the equations to solve and make use of quadratics to solve the problem.

If you have no knowledge of quadratics then you world need to look up some basic maths books or "google" for some simple questions involving simplifying and factorising quadratic expressions before solving equations and then attempt to do worded questions like this one.

We are talking about very basic algebra to get started, what you would do in first or second year secondary school.

For this question now that we have more or less derived the quadratic equation for this problem, the problem is really very simple and you should find lots of similar examples in nearly any basic maths book.
Guest
 

Re: Amount of Money Withdrawn

Postby Guest » Fri May 06, 2016 7:45 pm

Ok, thanks again.
Guest
 

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