A tank of volume 475 gallons is completely filled using pipes A and B

Plugging in numbers for this question wouldn't be a bad idea, we have 3 equations and 3 variables so we should at least know it is possible to work out all the variables. The idea for this strategy is to plug in a reasonable t and check if the 3 equations agree with each other on the solution for A and B. We would start from t =16 and work our way towards lower values as t must be less than 19.

To solve for t directly, we should approach the easiest equations first and eliminate the variables one by one. Set the time B was open as t, the efficiencies as A and B, the equations would be:

\(A(t + 5) + Bt = 475\)

\((A + B)*19 = 475\)

\(2*A*t = B*(t + 5)\)

The second equation gives \(A + B = 25\). Note we can write the first equation as \(t*(A + B) + 5A = 25t + 5A = 475\) which simplifies to \(5t + A = 95\). Finally we need to replace A in terms of t, we can use the 3rd equation to achieve that. Plug in \(A + B = 25\) as \(B = 25 - A\), the 3rd equation is:

\(2At = (25 - A)(t + 5)\)

\(2At + A(t + 5) = 25(t + 5)\)

\(A(2t + t + 5) = 25(t + 5)\)

\(A = 25\frac{t+5}{3t + 5}\)

Finally plug that back in \(5t + A = 95\):

\(5t + 25\frac{t+5}{3t + 5} = 95\), divide both sides by 5 and multiply by \(3t+5\).

\(t(3t + 5) + 5(t + 5) = 19(3t + 5)\)

\(3t^2 +10t - 57t - 14*5 = 0\)

\(3t^2 - 47t - 70 = 0\)

This is as far as I got, the answer from OP gives us A = 15, B = 10 and t = 16. Maybe I didn't understand the 3rd equation properly or the numbers are off.