Any GCSE maths teachers around please? Simultaneous equations...

[Celeriacacaca]

DS is very shaky on these and, having gone through some with him, I'm not sure I get them completely. Can someone please point me to a website or some examples to practice? He's getting questions such as 5x +3y=14
2x-y=10, solve simultaneously or

3 rulers and 4 pens cost £1.02.
2 rulers and 5 pens cost 96 pence
Calculate the cost of each ruler and each pen. He gets in this one he needs to multiply by 3 to get a common whatever it's called and can get the answer but is unsure about deciding when/where/what to multiply by.

Can anyone help please?

How about this video? corbettmaths.com/2013/03/05/simultaneous-equations-elimination-method/

Does he have access to mymaths or hegartymaths through the school?

If the video doesn’t help, there are a million other resources here for simultaneous equations, maybe one will click?

mrbartonmaths.com/topics/solving-equations/solving-simultaneous-equations/

I’d suggest watching a video with pen and paper, pausing it to see if he can think of the next step himself then checking. Once he’s feeling more confident he could try some worksheet questions. He must check his answers!

i don;t know if this will help you but in essence you want to get yourself to a position of only having one unknown thing. In your first example, this means making the equations have only x or only y. In the second example, you'd have only rulers or only pens. You do this by rearranging one of the equations.

So in your first example:

5x +3y = 14. Which means that 5x = 14-3y, which in turn means that x= (14-3y)/5.

Then you put this in place of the x in the other equation. So:
2x-y= 10 becomes [2* (14-3y)/5] - y = 10. Which means (28-6y)/5 -y=10. Times through by 5 to get rid of the fraction:

28-6y-5y = 50
Put your y's together: 28-11y = 50 > -11y = 50-28 > -11y=22 --> y = -2
Put that in the other equation: 2x - (-2) = 10 > 2x +2 = 10 > 2x=10-2 = 8 --> x=4

For your second example:
If r= rulers and p = pens then (1) 3r+4p=102, (2) 2r+5p = 96;
From equation 2: 2r = 96-5p --> r= (96-5p)/2
Subsititute this r into equation 1: 3*[(96-5p)/2] + 4p = 102
Times through by 2 to get rid of the fraction: 3*(96-5p) + 8p = 204
--> 288 - 15p + 8p = 204
-->288-204 = 15p-8p
-->84= 7p
--> p = 12
Pen costs 12p

Then in equation 2: 2r = 96-5p
--> 2r = 96-(5*12)
-> 2r = 96-60 = 36
--> r = 18
Ruler costs 18p

Of course they might teach it differently nowadays, so I hope this does not confuse.

Thanks all - perfect - really helpful. I was a bit lost and these are great.

Good idea to look at his my maths account giraffe as I'm sure that will have some too.

You want to have the same number of one of the unknowns in both equation so you can eliminate one unknown.

3 rulers and 4 pens cost £1.02 (eqn 1) (x2)
2 rulers and 5 pens cost £0.96 (eqn 2) (x3)

6 rulers and 8 pens cost £2.04 (eqn 3)
6 rulers and 15 pens cost £2.88 (eqn 4)

(eqn 4) minus (eqn 3) This way around so you get a positive answer for the number of pens.

(6-6) rulers and (15-8) pens cost (£2.88-£2.04)
0 rulers and 7 pens cost £0.84
so
1 pen costs £0.12

Substitute the cost of a pen in (eqn 1) or (eqn 2) to work out the cost of a ruler

Sorry, not a GCSE teacher but have studied maths to second year at university.