Please help, need to help a 10 year old with her homework

[mathbear]

I can’t get a clue of this question, could anyone please help? Thank you.

 

[Deleted member 4993]

I can’t get a clue of this question, could anyone please help? Thank you.

Please share the context of this problem.

What is being taught in the class - what is the topic? The problem can be solved using principle of simultaneous equation - but that may not be expected from 10 yr olds!

 

[Otis]

2B = 3∑
4∑ = 2B + ∂ +1
2∂ + 3∑ = 3B

Hi mathbear. Has your 10-year-old learned how to solve equations? Does she have experience substituting one expression for another? We can answer this exercise using such steps.

I'll use choice E, as an example:

∂ + 1 < ∑

Notice that inequality E contains these two symbols only: ∂, ∑

Which of the given equations contain those symbols? Not the first one, but the other two do:

4∑ = 2B + ∂ +1
2∂ + 3∑ = 3B

We would like to eliminate symbol B from either of those equations, in order to obtain an equation containing symbols ∂ and ∑ only . The easiest way is to make a substitution for 2B because they've already given us an expression for 2B in terms of symbol ∑. The first given equation tells us that 2B is the same amount as 3∑. Therefore, we may substitute the expression 3∑ for 2B anywhere it appears. Let's make that substitution in the second given equation:

4∑ = 2B + ∂ +1
becomes
4∑ = 3∑ + ∂ +1

Next, we would like to put this new equation into the same form as inequality E. That is, we would like to have the expression ∂ +1 by itself on one side of the equation. We can do that by subtracting 3∑ from each side.

4∑ – 3∑ = 3∑ + ∂ +1 – 3∑

Simplify:

∑ = ∂ + 1

Now, choice E tells us that amount ∂ +1 is less than amount ∑. That statement also means amount ∑ is greater than amount ∂ +1.

∑ > ∂ + 1

But we've already shown above (by substitution and solving for symbol ∑) that:

∑ = ∂ + 1

Therefore, inequality E is definitely not true. The expressions ∑ and ∂ + 1 are equal.

These are the sorts of steps that need to be done, in order to eliminate the given choices until we find the one that's true.

If your 10-year-old understands my example, then have her experiment solving equations for a symbol and making substitutions in other equations. Sometimes, we may need to make substitutions in two equations or solving each for a specific symbol, or make two substitutions in a single equation. The goal is to match the form of given choices, to see whether each is true or false.

If she needs more help, please have her post her efforts or ask specific questions. Thank you!



[imath]\;[/imath]

 

[The Highlander]

I can’t get a clue of this question, could anyone please help? Thank you.

Hi there,

I would like to offer another approach you might adopt.

If we number the equations thus:-

1. 2B̈ = 3∑​

2. 4∑ = 2B̈ + ∂ + 1​

3. 2∂ + 3∑ = 3B̈​

Now, since we know (from Equation 1) that 2B̈ is the same as 3∑ then we can replace 2B̈ in Equation 2 with 3∑ and we can also replace the 3∑ in Equation 3 with 2B̈.

So we may now re-write the equations as:-

1. 2B̈ = 3∑​

2. 4∑ = 3∑ + ∂ + 1​

3. 2∂ + 2B̈ = 3B̈​

Now, looking at the 'new' equations…

What does the new Equation 3 tell you about 2∂ ?

and what does the new Equation 2 tell you about ∂ +1?

They should tell you something about what (∂ +1) and 2∂ (and thence ∂) are equal to.

If you know what ∂, 2∂ and (∂ +1) are equal to that can help you to work out which statements are definitely not true (or could be True or False)!

For example, this knowledge will let you discount Statement E as definitely FALSE.
It also allows you to say that Statement B cannot be true either!
(Because B: 2∑ + B̈ < 4∂ may now be written as 2∂ +2 +2∂ <4∂ ie: 4∂ +2 < 4∂ which, clearly, can't be true)

Do you see why? ?

If you can follow this reasoning then you should be able to eliminate four of the five statements as not possibly true (or only maybe true) leaving just the one statement that is definitely TRUE. (It will also enable you to confirm that a particular statement is, definitely TRUE!) ?

I will leave it to you to figure out which ones are which but, if you get really stuck, let us know what you have managed to deduce and we will help you complete the process. ?

 

[Deleted member 4993]

now I feel like I need to go to school again

It is like Hotel California -

You can quit school anytime you want -

(but) you can never leave.

 

[Dr.Peterson]

now I feel like I need to go to school again

It's really just a system of equations in three variables: [math]2x=3y\\4y=2x+z+1\\2z+3y=3x[/math]
It can be solved by the usual methods, giving integer solutions, which can be put into each of the choices, and only one of them is true.

So as a problem for an algebra class, it is nothing special, once you get past the odd symbols. As a problem for a 10-year-old, or for an adult who has forgotten his algebra, ... well, you might not want to go to that school.

And if ancient hieroglyphics included modern numbers and operation symbols along with three symbols of their own, that would be amazing! So it's not very good history, either.

 

[Steven G]

1. 2B̈ = 3∑
2. 4∑ = 3∑ + ∂ + 1
3. 2∂ + 2B̈ = 3B̈

I would let ∑ =4 (an even number).

2B̈ = 3∑= 3*4=12, so B̈=6

4∑ = 3∑ + ∂ + 1
4*4 =3*4 + ∂ + 1
16=12 + ∂ + 1, so ∂ = 3

Which choice is true?

 

[Dr.Peterson]

1. 2B̈ = 3∑
2. 4∑ = 3∑ + ∂ + 1
3. 2∂ + 2B̈ = 3B̈

I would let ∑ =4 (an even number).

2B̈ = 3∑= 3*4=12, so B̈=6

4∑ = 3∑ + ∂ + 1
4*4 =3*4 + ∂ + 1
16=12 + ∂ + 1, so ∂ = 3

Which choice is true?

I don't think it's a good idea in solving a system of 3 equations in 3 unknowns, to just guess one of the variables. It just happens that 4 is the correct value; if not, your solutions would not satisfy the third equation, which you never used! (I hope you at least checked that it does.)

 

[Steven G]

Yes, I checked.