Sarah owns a car that runs on a mixture of gasoline and ethanol. she can buy fuels that have 85% ethanol or 25% ethanol. how much of each type of fuel should she buy if she wants to fill her 20 gallon tank with a mixture of fuel that contains 50% ethanol?

Denise needs to buy 25/3 gallons of 85% ethanol and 35/3 gallons of 25% ethanol to have 20 gallons of 50% ethanol.

x | y | 20 gal

use x and y because you don't know how much she needs.

0.85x | 0.25y | 20(0.5)

85% is 85/100 or 0.85, and you need that much of x, same goes for the 25% and 50% mixtures

so now you can make up 2 equations

1) x + y = 20

2) 0.85x + 0.25y= 10 (you get 10 when you multiply 20 by 0.5)

now you can solve for x or y using substitution. first rewrite 1) in terms of x or y:

x+ y= 20 > y= 20 - x

now you can substitute 20- x for y in the second equation..

0.85x + 0.25y= 10

0.85x + 0.25(20-x)= 10 distribute here..(0.25 * 20 and 0.25 * (-x) )

0.85x + 5 - 0.25x = 10 combine like terms

0.6x +5 = 10 move the 5 over to the other side

0.6x= 10 -5

0.6x = 5 divide both sides by 0.6

x= 25/3 or 8.3 now you know the amount of x so you can substitue this back into the first equation to find y.

0.85x + 0.25y= 10

0.85(25/3) +0.25y= 10

85/12 + 0.25y= 10

0.25y = 10- 85/12

0.25y= 35/12

y= 35/3 or 11.6

you can check by putting these values into the euations:

1) x+ y= 20

25/3 + 35/3 =20

20= 20 good so far

2) 0.85x + 0.25y= 10

0.85(25/3) + 0.25(35/3)=10

10 = 10 so our values for x and y work