Y6 maths question

[Robindrama]

7a-4b=2
2a +5b=19
How do they teach it now?
Thank you

I don't know if I've done it the way they teach it but here goes:
7a-4b +2a+5b =21 (adding the 2 equations)
9a +b =21 (add/subtract variables)
b=21-9a (now substitute into an equation)

7a -4(21-9a) = 2
7a -84 + 36a = 2 (multiplying out brackets)
43a = 2+84 = 86
a = 2 (put this in an equation)

2(2) + 5b = 19
5b = 19-4 = 15
b = 3

Y6 TA here, I would say the letter is in place of a value. A number right next to a letter means ×.
So 7×a - 4×b = 2
Last year most of our Year 6 were able to work out a and b using their tables knowledge and a bit of trial and error.
7x2 - 4x3 =2

a=2 b=3

2a + 5b =19
2×a + 5xb =19
Think about 5 and 2 times table, think of numbers that make 19
4 and 15
2x2 + 5x3 =19
a =2 b=3

I asked my yr6 child to do it the way they do it in school, trial and error. (not using algebra.)

He has done as:

7a - 4b = 2, meaning theres 2 difference between 7a and 4b.
He guessed a = 2 and b = 3, so 7 x 2 = 14, 4 x 3 = 12, 14 - 12 = 2, so it works.
Then he tested on a= 2, b= 3 to the second equation.
2a + 5b = 19.

(2 x 2 ) + ( 5 x 3) = 4 + 15 = 19.

So it's the right answer.
a = 2, b = 3.

I would agree y6 is trial and error.
My y9 DD has not yet done simultaneous equations.

That’s a very hard question for year 6 and I’d be surprised if the teacher had taught the year 6 algebra content yet. I’m still ploughing through mult and div fractions! We don’t teach them simultaneous equations but in a much simplier form they can use the same idea. So for example of the had 2b=10 and they also had 2a + 4b= 26 then they could substitute the 2b bit into the second expression.
For the one you give then trial and error is probably the way they would go. Parents forget how tricky the start of formal algebra can be to children as it seems so obvious to us. It can get far to hard very quickly!