percent of fruit juice

Call x the number of pints of the 40%  drink that are necessary.....then 140-x  is the number of pints necessary for the 60% drink.....and we have

.40x + .60(140 - x)  =.45(140)  simplify

.40x + 84 - .60x  = 63 

-.20x  = -21   divide both sides by -.20

x = 105 pints of the 40% juice

140 - x   =  140 - 105  = 35 pints of the 60% juice

The first type is 40% pure fruit juice, and the second type is 60% pure fruit juice. The company is attempting to produce a fruit drink that contains 45% pure fruit juice. How many pints of each of the two existing types of drink must be used to make 140 pints of a mixture that is 45% pure fruit juice?

Let the number of pints for 40% juice=x

Let the number of pints for 60% juice=y

(.4x  + .6y)=.45(x+y)

(.4x + .6y)=.45(140)=63

Solve the following system:
{0.4 x+0.6 y = 63
0.4 x+0.6 y = 0.45 (x+y)

0.45 (x+y) = +(0.45 x+0.45 y) = 0.45 x+0.45 y:
{0.4 x+0.6 y = 63
0.4 x+0.6 y = 0.45 x+0.45 y

In the first equation, look to solve for x:
{0.4 x+0.6 y = 63
0.4 x+0.6 y = 0.45 x+0.45 y

0.4 x+0.6 y = (2 x)/5+(3 y)/5:
(2 x)/5+(3 y)/5 = 63

Subtract (3 y)/5 from both sides:
{(2 x)/5 = -3/5 (y-105)
0.4 x+0.6 y = 0.45 x+0.45 y

Multiply both sides by 5/2:
{x = -3/2 (y-105)
0.4 x+0.6 y = 0.45 x+0.45 y

Substitute x = -3/2 (y-105) into the second equation:
{x = -3/2 (y-105)
0.6 y-0.6 (y-105) = 0.45 y-0.675 (y-105)

0.6 y-0.6 (y-105) = 0.6 y+(63.-0.6 y) = 63.-1.11022×10^-16 y:
{x = -3/2 (y-105)
63.-1.11022×10^-16 y = 0.45 y-0.675 (y-105)

0.45 y-0.675 (y-105) = 0.45 y+(70.875-0.675 y) = 70.875-0.225 y:
{x = -3/2 (y-105)
63.-1.11022×10^-16 y = 70.875-0.225 y

In the second equation, look to solve for y:
{x = -3/2 (y-105)
63.-1.11022×10^-16 y = 70.875-0.225 y

Subtract 63.-0.225 y from both sides:
{x = -3/2 (y-105)
0.225 y = 7.875

Divide both sides by 0.225:
{x = -3/2 (y-105)
y = 35.

Substitute y = 35. into the first equation:
Answer: | x = 105.                   andy=35
 

CPhill: You are a master manipulator of these "word problems". Congrats!.