Let C = the capacity of the tank. Let r1 and r2 be the initial flow rates through the first and second pipes respectively. Let r1_new and r2_new be the new flow rates. Let t1 and t2 be the initial time it takes to fill the tank using the initial flow rates of the first and second pipes respectively.
Using the formula Capacity = Rate * Time (and its rearrangements), we can form the following six equations:
Equation 1: r1 + r2 = 3C/20
Equations 2 and 3: r1 = C/t1 ; r2 = C/t2
Equations 4 and 5: r1_new = C/(t1 - 1) ; r2_new = C/(t2 + 2)
Equation 6: r1_new + r2_new = C/7.
I now substitute Eqs. 2 and 3 into 1, and Eqs. 4 and 5 into 6, to get the following:
}+\frac{C}{(t2+2)}=\frac{C}{7})
Now I factor out C from both equations (C > 0, or else the question makes no sense). We arrive at:
}+\frac{1}{(t2+2)}=\frac{1}{7})
Our goal is therefore to work out the value of t1 and t2, so we can work out (t1-1) and (t2+2), which is the ultimate aim.
Algebraically manipulating the fractions, we get the following:

;
(t1+2)}=\frac{1}{7} \implies 7t1+7t2+7=t1t2+2t1-t2-2 \implies 5t1+8t2+9=t1t2 \implies 15t1+24t2+27=3t1t2)
. I now treat this as a system of (non-standard) simultaneous equations, and try to solve. We get:

. But notice that we previously had

Therefore, we may subtract these two equations to get
=36 \implies t2=\frac{36}{t1-12})
We can substitute this back into 5t1-4t2=27 to get
-144=27(t1-12)\implies5(t1)^{2}-60t1-144=27t1-324\implies 5(t1)^{2}-87t1+180=0)
This quadratic can be factorised to give
(t1-15)=0\implies t1\in\{\frac{12}{5},15\})
. If you substitute these values back into 5t1-4t2=27, you will get the corresponding solutions

Of course, we can't have a negative time value, so it follows that

. Those are the initial times to fill the tank for each pipe.
To get the new times, we simply evaluate t1-1 and t2+2, to get new times of 14 minutes for both pipes! Therefore, under the new flow rates, each pipe will take 14 minutes to fill the tank!