THREE LINEAR EQUATIONS WITH THREE VARIABLES

Question

A chemist has three different acid solutions. The first acid solution contains 15% acid, the second contains 25%,
and the third contains 70%. He wants to use all three solutions to obtain a mixture of 40 liters containing 45%
acid, using 2 times as much of the 70% solution as the 25% solution. How many liters of each solution should be used?

How to set up as three linear equations to find the answer?

Answer 1

set up 2 equations

the first equation...

15%*x+25%*y+70%*2y=45%*40

0.15x+0.25y+1.4y=0.45*40

0.15x+1.65y=18

multiply all terms by 100

15x+165y=1800

divide all terms by 15

x+11y=120

the second equation...

x+y+2y=40

x+3y=40

x=40-3y

substitute into the first equation x+11y=120

40-3y+11y=120

40+8y=120

8y=120-40

8y=80

y=80/8

y=10 liters

x+3*10=40

x+30=40

x=40-30

x=10 liters

10 liters of 15%, 10 liters of 25%, and 20 liters of 70%

check:

10*15%=1.5

10*25%=2.5

20*70%=14

1.5+2.5+14=18

18/40=9/20=45/100=45%

45%*40=18