The reason my method is not traditionally taught to beginners is because it requires an algebraic technique that is a bit complex. But most of the students I have tutored face to face have found it an intuitive way to solve all word problems once they learn that technique. It simplifies the thought process by using more complex but mechanical algebra.
Here goes.
Step 1: Identification. We ask ourselves what relevant numbers are unknown, and we assign in writing (so we don’t forget) a letter to each. So in this problem, I suggested
[MATH]f = \text {number of marbles taken by first brother.}[/MATH]
[MATH]s = \text {number of marbles taken by second brother.} [/MATH]
[MATH]t = \text {total number of marbles.}[/MATH]
The letters are of course arbitrary. I chose f for first, s for second, and t for total just for mnemonic reasons.
For most school problems, identifying what you need to know but don’t yet know is usually quite easy. (In the real world, this may not be so easy.)
Understand step 1?
The result of step 1 is almost always that we have more than one unknown number. Let’s say the number of unknowns is n > 1. We can solve such a mess if we can find n equations that relate the unknowns together.
Step 2. Translation. A word problem gives us information about the unknown numbers, but it does so in English. We need to translate the English into algebraic equations, and we need n (three in this case) equations. Two are super easy.
After both brothers have taken marbles, 10 are left. This translates algebraically into
[MATH]t - f - s = 10.[/MATH]
The first brother takes 4/7ths of the marbles. This translates algebraically into
[MATH]f = \dfrac{4}{7} * t = \dfrac{4t}{7}.[/MATH]
Easy peasy. The third equation is not so easy. The second brother took 7/9ths of what was left. How do we translate that. Well what was left is t - f so
[MATH]s = \dfrac{7}{9} * (t - f).[/MATH]
We now have three equations. Finding the equations may take some thought and care.
Understand what we do in step 2? Does it make sense?
Step 3: Solve algebraically.
Here is why this method is not taught to beginners. You have what is called a system of equations. They are more complicated to solve than a single equation. But it is not that complicated if all the equations are linear. What we do is to find out what one of the unknowns is in terms of the others and then use that information to reduce the system to n - 1 equations in n - 1 unknowns. We repeat that process until we are down to one equation in one unknown and then solve that. Finding the easiest way to do that requires experience, but you can just do it by brute force.
[MATH]f = \dfrac{4t}{7} \text { and } s = \dfrac{7}{9} * (t - f) \implies[/MATH]
[MATH]s = \dfrac{7}{9} * \left ( t - \dfrac{4t}{7} \right ).[/MATH]
Does that make sense?
Let's simplify that.
[MATH]s = \dfrac{7}{9} * \left ( t - \dfrac{4t}{7} \right ) = \dfrac{7}{9} * \left ( \dfrac{7t}{7} - \dfrac{4t}{7} \right ) = \dfrac{\cancel 7}{3 * \cancel 3} * \dfrac{\cancel 3t}{\cancel 7} = \dfrac{t}{3}.[/MATH]
We now reduce the system of three equations to just two equations in two unknowns.
[MATH]t - f - s = 10 \implies t - f - \dfrac{t}{3} = 10 \text { because } s = \dfrac{t}{3}.[/MATH]
But we already know what f is in terms of t.
[MATH]t - f - s = 10 \implies t - \dfrac{4t}{7} - \dfrac{t}{3} = 10 \text { because } f = \dfrac{4t}{7}.[/MATH]
We now have a single equation in one unknown and can solve that easily.
[MATH]t - \dfrac{4t}{7} - \dfrac{t}{3} = 10 \implies 21t - 12t - 7t = 210 \implies 2t = 210 \implies t = 105.[/MATH]
It is now a snap to find f and s.
[MATH]f = \dfrac{4 * 105}{7} = 4 * 15 = 60,\ s = \dfrac{105}{3} = 35.[/MATH]
Step 4: Checking. It is easy to make mistakes. The fourth step is to check that you did not make any.
[MATH]105 - 60 - 35 = 45 - 35 = 10. \checkmark.[/MATH]
Now as I said, this method puts far less strain on the imagination when solving word problems, but it does require learning how to deal with systems of linear equations. There is undoubtedly a later part of your book that tells you how to do exactly that. We can help you do the exercises that teach you the mechanics of solving systems of linear equations, and you will then find word problems much easier to deal with.
