I have two double-sided cards, and using both cards I can make these values: 7,10,-1,-4.?

Question

What integers are on each side of the two cards?

Explain how you worked out your answer.

Sorry I am trying to help my daughter with this but I have brain drain.

Can anybody please help.

Thank you.

My daughter is only ten years old. Is there a more simple way to explain to her?

Answer 1

This is not an easy one, so you should not feel bad that you could not work it out yourself.

How do the values pair up into two pairs with the same difference?

Answer 1: (7, 10) and (-4, -1), difference 3.
Answer 2: (-4, 7) and (-1, 10), difference 11.

So we deduce that one card has the integers A and A+3, which means that flipping it alters the total value showing by 3; while the other card has the integers B and B+11, which means that flipping it alters the total value showing by 11.

If the value showing is -4 and flipping the B card changes it to 7, then the A card must be showing A, not A+3. When it shows A+3, the total is -1, and flipping the B card changes it to 10.

So the equations are A + B = -4, A + (B+11) = 7, (A+3) + B = -1, and (A+3) + (B+11) = 10. But all of these just reduce to the same equation A + B = -4, and there are any number of possible solutions. For example :-

With one card having the integers (-2, 1) and the other (-2, 9) you can make -4, -1, 7 and 10.

With one card having the integers (-20, -17) and the other (16, 27) you can make -4, -1, 7 and 10.

With one card having the integers (-53, -42) and the other (49, 52) you can make -4, -1, 7 and 10.

I don't know what your daughter can tell her teacher, except show her three or four different solutions which all work. As long as the smaller integers on each of the two cards add up to -4, and the larger integer is 3 greater on one card and 11 greater on the other, those cards will make the values given in the problem.

Answer 2

If you use the simultaneous equations that have been described above, you should get the following answers:

Card 1 has sides of -7 and 4.
Card 2 has sides of 3 and 6.

And as a result: -7 + 3 = -4, -7 + 6 = -1, 4 + 3 = 7, and 4 + 6 = 10.

I just kinda "Trial and error"ed it, which probably still isn't the best way to explain it. :)

Answer 3

lets assign some values, Card A has two sides, 1 and 2. Card B has two sides 1 and 2
A1+B1=10
A1+B2=7
A2+B1=-1
A2+B2=-4

Right off you should be able to see that A2 needs to be a negative number less than -4. Lets assign -5
-5+B2 = -4 .... B2= 1
-5+B1 = -1 .... B1= 4
Now plug in for A1
A1 + 4 = 10 ..... A1= 6
A1 + 1 = 7 ....... A1 = 6

I think you will find that a number of other combinations will also work. Plug them in and try them

Answer 4

they're simultaneous equations. start with

assume one card has A on one side and B on the other
and the 2nd has x on one and y on the other
assign each equation a value and go to work!
e.g.
A+x=7
B+x=10
A+y=-1
B+y=-4

ill take a little while theres a lot to do

Answer 5

The first card has values of w and x. The second has values y and z.

w+y=7
w+z=10
x+y=-1
x+z=-4

Four equations, four unknowns.