Can anyone do my maths homework?

[Elisheva]

Not actually my homework. It’s from my niece who is in year 7. I don’t think it is possible, but I’m not very good at maths!

Could there be negative numbers involved?

For Q5 you have
x+y=7
x-y=4

Adding gives 2x=11 so x=5.5
But two lots of 5.5 isn’t 10 so the bottom equals doesn’t work.
y is 1.5 but two lots of 1.5 isn’t 8.

Unless I’ve misunderstood the question?

Oh, sorry, completely misunderstood, the numbers in the circles aren’t the same, ignore me!

There could be negative numbers, but that doesn’t seem to help!

3+4=7

Nope. Number 5 must have a mistake.

From top left clockwise 2.5, 4.5, 3.5, 7.5

Q6:
10.5 + 1.5
5.5 + 3.5

Thank you! I knew mumsnet would know 😊

I agree with green tulips

Second one 10.5, 1.5, 3.5, 5.5

a+b = 7
x-y = 4
a+x = 10
b+ y = 8

So adding equations 2 and 4 you get

x + b = 12

Subtract the first equation from this

x - a = 5

Add this to the third equation

2x = 15

So x = 7.5
a = 2.5
b = 4.5
y = 3.5

Does that work?

So the numbers are

2.5. 4.5
7.5. 3.5

I solved #5, I think, like this:

I used a, b, c, d as the blanks. The equations become
a + b = 7
c - d = 4
a + c = 10
b + d = 8

adding
c - d = 4
b + d = 8

c + b = 12

<span class="underline">then, subtracting</span>
c + b = 12 (from the calculation)
a + b = 7 (from the problem)
~~~~~~
c - a = 5

<span class="underline">then, adding</span>
a + c = 10 (from the problem)
c - a = 5 (from the calculation)
~~~~~~
2c = 15
c = 7.5

and the rest is easy.

The longest post I've ever made on Mumsnet.  I hope I didn't make any typos.

You’ve got 4 sets of equations for each one. If top left is x, top right is y, bottom left is a and bottom right is b, you then have (for question 5):
x+y=7
a-b=4
x+a=10
y+b=8

You can then rearrange and substitute the equations until you figure out the value of one of the letters. You can then figure out the rest of the numbers easily.

Without showing working, I got x=2.5, y=4.5, a=7.5 and b=3.5 for question 5, and x=10.5, y=1.5, a=5.5 and b=3.5 for question 6.

Has she been doing algebra at school? This is how I assume she should be working them all out. These are slightly trickier as they aren’t whole numbers.

Okay, here's the formal way of solving question 5 (I suspect your niece is actually being expected to use trial-and-error).

Labeling the missing numbers in the circles a, b (top row, L to right) and c, d (bottom row, L to right), we end up with 4 equations in 4 unknowns.

so

a+b=7 (eqn1)
c-d=4 (eqn2)
a+c=10 (eqn3)
b+d=8 (eqn4)

subtract eqn1 from eqn 3 (to get rid of a)
c-b=3 (eqn5)

add eqn2 and eqn 4
c+ b = 12 (eqn6)

add eqn5 and eqn6
2c=15 i.e. c=7.5

substitute c back into eqn 3
a=2.5

substitute c back into 2
d = 3.5

substitute a back into 1
b = 4.5

So the answer looks like this:

2.5 + 4.5 = 7

•         +
  

7.5 - 3.5 = 4

= =
10 8

(I hope your niece has been taught about simultaneous equations, because this would be a bit of a bugger to solve by trial-and-error.)

Lots of cross-posts while we all wrote it down!

Do you want the next question doing as well, or do you want to have a go at it now you know the method?

And realjudas gets 0 for not showing her working. But yes simultaneous equations is the way..... Seems tricky for Y7, but that was a very long time ago for me so I don't really remember...

Q6

You need to give each circle a different letter so that you end up with four equations

So
a+b =12
a-x = 5
b+y=5
x+y=9

Then you express a and b in terms of x and y

Looking at a first

a=12-b
Substituting this into a-x=5 gives 12-b-x=5 which simplifies to x=7-b

Next look at b

b=12-a
Substituting this into b+y=5 gives 12-a+y=5 which simplifies to y=a-7

Finally you gather your similar terms

You already have a+b=12 but now have what they are equivalent to in terms of x and y so x+y=9 becomes 7-b+a-7=9 which simplifies to a-b=9

So you are left with

a+b=12
a-b=9

Gather/ cancel your like terms and you get a+b+a-b=21 which simplifies to 2a=21, therefore a=10.5. Once you have this you can solve for the other letters, b=1.5, x=5.5, y=3.5

I knew the answer just to look at q5 it but will look at the algebra.

I have a perhaps simpler method, with fewer letters.
Q5.

If the TopLeft = x
then top right is 7-x
and also bottom left is 10-x

For right hand column that means bottom right = x +1

but for bottom row you need
10-x-(x+1) = 4
10-x-x-1=4
5=2x
x=2.5

then substitute in to the other circles giving 2.5, 4.5, 7.5, 3.5

Oh, neat TeenTimesTwo.

I always like a terse, elegant solution.

Fermat So do I.
I always say that mathematicians are inherently lazy so like to do things simply if at all possible; like spotting when to simplify fractions before multiplying - that kind of thing.