11 Plus Exams Forum

Hi all. Unfortunately the last time I did any algebra was many, many years ago and I don't remember being any good at it then! I wonder if I could get some help?

In the following question does it matter which side you start on?

36k+15 = 50k-27

I'm sure this is pretty basic stuff to most of you but I can't get my head around it!

Do I need to cancel out the - number first?

Confused!

I'm sure a maths teacher will be on to tell you how to do it soon but this is how I would do it:

36k + 15 = 50K - 27

Add 27 to both sides

36k + 15 + 27 = 50k

36k + 42 = 50k

Minus 36k from both sides

42 = 50k - 36k

42 = 14k

3 = k

Does that help?

Great thanks!

Yes it does help as that was how I was thinking I would do it.

Would it change do you know if the minus number was on the other side? So ..

19u-36 = 45u+16

Would I then start by adding 36?

thescribe wrote:Great thanks!

Yes it does help as that was how I was thinking I would do it.

Would it change do you know if the minus number was on the other side? So ..

19u-36 = 45u+16

Would I then start by adding 36?

As a maths teacher I would always look at the unknown term first and decide which side of the equation had most. In this case the RHS has more u so I will need to collect u on that side and constants on the other.

-16 from both sides (the balance method)

19u - 36 - 16 = 45u

so -19u from both sides

etc.

What year group is this for?

What are you doing when you solve an equation? [More important than the 'method' is understanding the meaning of the answer]

Thank you!

He's in Y5 but this is something his tutor has given him for grammar prep. We thought he had understood it but he says he got it wrong and unfortunately my son got quite upset during the session and completely forgot what he told him about what had been done wrong etc.

This is how he did it:

19u + 36 = 45u -16

-36 on both sides

19u = 45u-52

-45u on both sides

-26u/-26 = -52/-26

u = 2

Its all over my head I'm afraid

My dd is in year 7 and did this earlier this term. She didn't do it at all for her grammar preparation. maybe you need it where you are but it might be worth finding out?

Yes it looks too hard for GS prep ... which is why I asked.

What you are doing when you solve a linear equation in one unknown is fidning the ONE value of the unknown that makes the equation true.

Having now "officially" retired from posting on this site, my DC having accepted their offered 2015 GS place, I'm just calling in briefly to show the straightforward way I was taught (many more moons ago than you might imagine) to do this kind of problem.

36k + 15 = 50k -27

rewrite as: + 36k +15 = + 50k - 27

move the + 36k and the - 27 across when they then become -36k and + 27 respectively and thus rewrite as: + 15 + 27 = + 50k - 36k

which is: 42 = 14k (i.e. 14 x k)

and therefore: 42/14 = k

hence: k = 3

Simples, as they say these days!

thanks for all the replies.

I must admit I think it's hard, especially as they've not even done algebra at school yet and he's one of the youngest in the class too. (not sure that matters!)

But, his tutor only tutors for Grammar prep and has done for years and he says it's necessary so now I'm wondering if he's pushing too hard!

Please ignore equilibrio's method - swapping sides and swapping signs does NOT make sense!

When he suggests this step

which is: 42 = 14k (i.e. 14 x k)

and therefore: 42/14 = k

note the 14 is 'swapping sides' but does not change sign - this is why this method is NOT taught now.

The balance (or inverse operation) method is clear and needs no tricks.