If [tex]A(t)[/tex] is the amount of salt in the tank at time [tex]t[/tex], then the rate at which the amount of salt in the tank changes is given by
[tex]\dfrac{\mathrm dA(t)}{\mathrm dt}=\dfrac{4\text{ lbs}}{1\text{ gal}}\dfrac{6\text{ gal}}{1\text{ min}}-\dfrac{A(t)\text{ lbs}}{600\text{ gal}}\dfrac{6\text{ gal}}{1\text{ min}}[/tex]
[tex]\dfrac{\mathrm dA}{\mathrm dt}=24\dfrac{\text{lb}}{\text{min}}-\dfrac{A(t)}{100}\dfrac{\text{lb}}{\text{min}}[/tex]
Let's drop the units for now. We have
[tex]\dfrac{\mathrm dA(t)}{\mathrm dt}+\dfrac{A(t)}{100}=24[/tex]
[tex]e^{t/100}\dfrac{\mathrm dA(t)}{\mathrm dt}+e^{t/100}\dfrac{A(t)}{100}=24e^{t/100}[/tex]
[tex]\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/100}A(t)\right]=24e^{t/100}[/tex]
[tex]e^{t/100}A(t)=\displaystyle24\int e^{t/100}\,\mathrm dt[/tex]
[tex]e^{t/100}A(t)=2400e^{t/100}+C[/tex]
[tex]A(t)=2400+Ce^{-t/100}[/tex]
We're given that the water is pure at the start, so [tex]A(0)=0[/tex], giving
[tex]A(0)=0=2400+Ce^{-0/100}\implies C=-2400[/tex]
So the amount of salt in the tank (in lbs) at time [tex]t[/tex] is
[tex]A(t)=2400\left(1-e^{-t/100}\right)[/tex]
