Question 955337: Please Help me work this out step by step... I Haven't done a mixture problem with so many variables before. I appreciate all the help I can get
Ethonal and Toluene are two examples of additives within your gasoline.
Type A gasoline is 5% Ethanol and 10% Toluene
Type B gasoline is 15% Ethanol and 8% Toluene
Type C gasoline is 10% Ethanol and 4% Toluene
A special Type D gasoline is 8.05% Ethanol and 6.84% Toluene. Precisely blending the correct quantities of Type A, B, and C together can create type D.
How much of each type should be blended together to create 100 gallons of Type D gasoline.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! let a = number of gallons of type A gasoline.
let b = number of gallons of type B gasoline.
let c = number of gallons of type C gasoline.
let d = number of gallons of type D gasoline.
D = 100 gallons.
you have 3 equations that need to be solved simultaneously.
they are:
a + b + c = 100
this equation tells you that the number of gallons of type A and type B and type C must be equal to the number of gallons of type D which is equal to 100 gallons.
.05a + .15b + .10c = .0805d = .0805 * 100 = 8.05
this equation tells you that the number of gallons of ethanol in type A and type B and type C gasoline must be equal to the number of gallons of ethanol in type D gasoline.
.10a + .08b + .04c = .0684d = .0684 * 100 = 6.84
this equation tells you that the number of gallons of toluene in type A and type B and type C gasoline must be equal to the number of gallons of toluene in type D gasoline.
the 3 equations are:
a + b + c = 100
.05a + .15b + .10c = 8.05
.10a + .08b + .04c = 6.84
when you solve these equations simultaneously, you will get:
a = 44
b = 5
c = 51
you will need 44 gallons of type A gasoline and 5 gallons of type B gasoline and 51 gallons of type C gasoline to make 100 gallons of type D gasoline.
solving this system of equations by elimination can be done as follows:
start with:
a + b + c = 100 (equation 1)
.05a + .15b + .10c = 8.05 (equation 2)
.10a + .08b + .04c = 6.84 (equation 3)
work with equations 1 and 2 to start.
multiply both sides of equation 1 by .05 to get:
.05a + .05b + .05c = 5 (equation 4)
.05a + .15b + .10c = 8.05 (equation 2)
subtract equation 4 from equation 2 to get:
.10b + .05c = 3.05 (equation 5)
now work with equation 1 and equation 3.
multiply both sides of equation 1 by .10 to get:
.10a + .10b + .10c = 10 (equation 6)
.10a + .08b + .04c = 6.84 (equation 3)
subtract equation 3 from equation 6 to get:
.02b + .06c = 3.16 (equation 7)
you now have 2 equations in 2 unknowns.
they are:
.10b + .05c = 3.05 (equation 5)
.02b + .06c = 3.16 (equation 7)
multiply both sides of equation 7 by 5 to get:
.10b + .05c = 3.05 (equation 5)
.10b + .30c + 15.8 (equation 8)
subtract equation 5 from equation 8 to get:
.25c = 12.75
solve for c to get:
c = 51
now that you have c = 51, you can go back to equation 5 or equation 8 and solve for b.
we'll use equation 5.
you get:
.10b + .05c = 3.05 (equation 5) becomes:
.10b + .05(51) = 3.05
solve for b to get:
.10b = 3.05 - .05(51) = .5
solve for b to get b = 5
you now have c = 51 and b = 5
go back to any of the original equations and solve for a.
we'll use equation 1.
you will get:
a + b + c = 100 (equation 1) becomes:
a + 5 + 51 = 100 which becomes:
a + 56 = 100
solve for a to get a = 44.
you now have:
a = 44
b = 5
c = 51
go back to your original equations and solve them using those values to see that all the equations are true.
they are, so the solution is correct.
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