SOLUTION: A 6.0% solution and a 13.0% solution of a drug are added to 200 mL of a 24.0% solution to make 1500 mL of a 12.0% solution. The following equations relate to the milliliters of th

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: A 6.0% solution and a 13.0% solution of a drug are added to 200 mL of a 24.0% solution to make 1500 mL of a 12.0% solution. The following equations relate to the milliliters of th      Log On

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Question 788371: A 6.0% solution and a 13.0% solution of a drug are added to 200 mL of a 24.0% solution to make 1500 mL of a 12.0% solution. The following equations relate to the milliliters of the added solutions. The number of milliliters of the added solution are represented by x and y, respectively. Find x and y (to three significant digits) x+y+200= 1500 .06x+.12y+.24(200)=.12(1500)
can someone show me how to do this with steps

Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
we are given two equations in two unknowns
x+y+200= 1500
.06x+.12y+.24(200)=.12(1500)
subtract 200 from both sides of the = in the first equation
x+y = 1300 and
y = 1300 - x
now multiply percentages in second equation
.06x +.12y +48 = 180
subtract 48 from both sides of =
.06x +.12y = 132
now substitute 1300-x for y
.06x +.12*(1300-x) = 132
.06x +156 -.12x = 132
subtract 132 from both sides of =
.06x +24 -.12x =0
.06x = 24
x = 400 mL
400 +y = 1300
y = 900 mL
now
.06 * 400 = 24 mL of 6% solution
.12 * 900 = 108 mL of 12% solution, checking
24 +108 +48 = 180
180 = 180