Answer:
Wyzant
ALGEBRA WORD PROBLEM
Dorian S. asked • 06/28/17
THREE LINEAR EQUATIONS WITH THREE VARIABLES
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?
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Arthur D. answered • 06/28/17
TUTOR 4.9 (67)
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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
hope it helps
please mark brainliest
If you divide decimals you have to bring up the decimal point but if you divide whole numbers you dont have any decimal points so you just divide the numbers. Sorry if i didnt help i just wanted to help.
Answer:
X=132-13A/A, A cannot equal 0
A=132/X+13, X cannot equal -13