Question 472969
Let x = amount of 30% solution, y = amount of 60% solution (both amounts in mL)


Since the chemist wants 300mL, this means that {{{x+y=300}}}. Solve for y to get {{{y=300-x}}}



Because "A chemist has a 30% acid solution and a 60% solution already prepared" and the chemist wants "to form 300 mL of a 50% solution", we know that {{{0.3x+0.6y=0.5(300)}}}. Next, multiply everything by 10 to get every number to be a whole number. So 



{{{10(0.3x)+10(0.6y)=10*0.5(300)}}}



{{{3x+6y=1500}}}



Now plug in {{{y=300-x}}} and solve for x


{{{3x+6y=1500}}}



{{{3x+6(300-x)=1500}}}



{{{3x+1800-6x=1500}}} Distribute.



{{{-3x+1800=1500}}} Combine like terms on the left side.



{{{-3x=1500-1800}}} Subtract {{{1800}}} from both sides.



{{{-3x=-300}}} Combine like terms on the right side.



{{{x=(-300)/(-3)}}} Divide both sides by {{{-3}}} to isolate {{{x}}}.



{{{x=100}}} Reduce.



Now go back to {{{y=300-x}}} and use that to find the value of y.



{{{y=300-x}}}



{{{y=300-100}}}



{{{y=200}}}



So {{{x=100}}} and {{{y=200}}}



This means that 100 mL of 30% acid solution and 200 mL of 60% acid solution is needed to mix to get 300 mL of 50% acid solution.