Very elementary algebra problem

Hi,

I am introducing algebra to my 6 year old.

We are solving problems of the type: 10 - x = 50
It does not appear that I can use the Solve function because it will reply with "No real roots". Of course x = -40 (a negative number)
Can I use SMath for these types of equations.

Much Thanks
Ron (dad) and Rachel

Hello Ron,

If you use "solve" there are few things one should take care of.
- This functions is a numerical function and can have two or four arguments.
- If you use the two arguments version here is your example:
[MATH=eng]solve(10-x≡50,x)=-40[/MATH]
As you can see, there is the right answer. The "No real roots" message appears when Tools|Options|Calculation and the default Roots(range) option does not consist the results. Check this option, please. If your range is, say [-20,20] then this message will appear because -40 is outside the roots range search. If you are explaining this to 6 year old child, I suppose that will confuse him - therefore check the range and change it properly.

- You can use the four argument solve which will ignore the default range and use the given one as the third and fourth argument, like this:
[MATH=eng]solve(10-x≡50,x,-50,50)=-40[/MATH]
However, I am not sure that the 6 year old child will understand this.

Regards,
Radovan

Radovan,

Hmm.. - Well that Option dialog box solved this problem, and I certainly want to use the 2 argument "Solve" function rather than the 4 argument choice with my 6 year old daughter, however... it looks like I have to keep the range between the lower limit to upper limit within 200.
Another way of stating this is, if I set the range in the option dialog box to [-100, 100], then this is fine.
If I set the range to [-100, 101], then SMath cannot solve this.... and than I have to use the 4 argument "solve" function - am I correct?

Much much Thanks
Ron

You are welcome Ron

Wrote

If I set the range to [-100, 101], then SMath cannot solve this.... and than I have to use the 4 argument "solve" function - am I correct?

Here it is ... you found it as well. That was mentioned few times here on the Forum some time ago. I can not tell you what is so strange in [100,101] interval. There might be some "glitch" in the solve function numerical algorithm. The right answer can give you the creator of that function. Maybe Andrey, the SMath creator, can give you the right answer.

Regards,
Radovan

Hello Ron
Mr. Omorr had given a good answer for it .
The equation :
10-x=50
This is a simple equation but the problems and the way it should be solved is the best way .So u should follow him or just follow this link algebra problem solver

how to Simplifying algebraic equations ??

I saw the question below in the 12th edition of the official study guide. I need further explanation of why it was solved the way it was solved in the official guide.

If x,y and k are positive numbers such that (x/x+y)(10) + (y/x+y)(20) = k and if x < y, which of the following could be the value of k?
(A) 10
( 12
(C) 15
(D) 18
(E) 30

Wrote

how to Simplifying algebraic equations ??

I saw the question below in the 12th edition of the official study guide. I need further explanation of why it was solved the way it was solved in the official guide.

If x,y and k are positive numbers such that (x/x+y)(10) + (y/x+y)(20) = k and if x < y, which of the following could be the value of k?
(A) 10
( 12
(C) 15
(D) 18
(E) 30

(10x + 20y)/(x+y) = k

10x + 20y = kx + ky

20y - ky = kx - 10x

y(20 - k) = x(k - 10)

(20 - k)/(k - 10) = x/y

and since 0 < x < y, then 0 < x/y < 1, and it must be that 0 < (20 - k) / (k - 10) < 1. From the answer choices we can be sure k - 10 isn't negative, so we can multiply through this inequality by k-10 to find that 0 < 20 - k < k - 10, or that 15 < k < 20.
So you can solve also Factor, factorization, trinomial, polynomial, binom .