Question 955337
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.