A little bit more general for all word problems could be something like this. I use this for my students:

1) Understand what the problem is saying - I've seen students see words and simply turn off. They don't even try to read the problem and simply try to understand what is going on. As in, here, things like:

- we have a tank filling up

- two pipes filling it up, one small and one large

- together, they fill it up in 6 hours

- along, the small pipe takes 2 hours longer than the large pipe

- find out how long the large pipe would take alone.

Forget the math, just understanding what the problem is talking about is difficult enough.

2) Set up the problem - once you understand the problem, it is a little easier to understand how to set it up. Like here, what can be difficult to understand:

the rate of the small pipe + the rate of the large pipe = the rate of the pipes together

Either understanding that "formula" is difficult or how the rate of the large pipe is 1/x and the small pipe is 1/x+2.

3) Solve for x

4) Check - you can always check every math problem. Here, you could try plugging in the solution for x and see of each side is equal to each other (at least approx, given decimals and round off error).

Since the problem only provides the time the tank is beiong filled by the two pipes together, you can use the following notations for the time each pipe fills the tank.

The larg pipe fills the tank in `x` hours and the small pipe fills the tank in `x + 2` hours.

Hence, the large pipe fills `1/x` of the tank in 1 hour and the small pipe fills `1/(x + 2)` of the tank in 1 hour.

Together, the large pipe and the small pipe fill 1/6 of the tank in 1 hour.

You may set up the following equation, such that:

`1/x + 1/(x + 2) = 1/6`

You need to bring the fractions to a common denominator, such that:

`6(x+2)/(6(x(x+2))) + (6x)/(6(x(x+2))) = (x(x+2))/(6(x(x+2)))`

`6(x + 2) + 6x = x(x + 2)`

`6x + 12 + 6x = x^2 + 2x => x^2 + 2x - 12x - 12 = 0`

`x^2 - 10x - 12 = 0 `

You should use quadratic equation, such that:

`x_(1,2) = (10 +- sqrt(100 + 48))/2 => x_(1,2) = (10 +- sqrt148)/2`

`x_(1,2) = (10 +- 2sqrt37)/2 => x_(1,2) = 5+-sqrt37`

You need to reject the negative value since x represents the number of hours, hence `x = 5+sqrt37 => x ~~ 11` hours.

**Hence, evaluating the time needed by the large pipe to fill the tank yields `x ~~ 11` hours.**