Question 793779
Best to view this as a mixture problem and focus on the concentration of milk in the two vessels.  Do as normal fractions if this is most comfortable.

vessel, 4/7 milk.
vessel other, 9/10 milk.
Want 3/5 milk.


We do not yet know how much from the 4/7 vessel nor from the 9/10 vessel, but we can assume any number of parts total in the final mixture.  


Try this:
x parts of the 4/7 milk
y parts of the 9/10 milk
{{{x+y=700}}} parts, chosen because of denominators 7 and 10.


Accounting for concentration of milk,
{{{((4/7)x+(9/10)y)/700=3/5}}}
Simplify this rational equation so it is in a neater linear equation format.  This and the 700 Sum equation will be your system to solve for x and y.





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{{{(4/7)x+(9/10)y=(700*3)/5}}}
{{{(4/7)x+9/10)y=140*3}}}
{{{70*(4/7)x+70(9/10)y=70*140*3}}}
{{{40x+63y=7*14*3*100}}}
{{{40x+63y=29400}}}---------one equation for the system
Try substituting with y=700-x.
{{{40x+63(700-x)=29400}}}
{{{40x+63*700-63x=29400}}}
{{{-23x=29400-63*700}}}
{{{23x=63*700-29400=14700
{{{23x=14700}}}
{{{x=14700/23}}}=
{{{x=639}}}
I'm getting very close to 639, but may have a mistake somewhere.
{{{y=61}}}, {{{x=639.}}}
Seems very near to choice (b).