SOLUTION: Solving Linear Systems of equations
The problem: one is 5% acid and one is 6.5% acid . You want to make 200ml of vinegar,with 6% acid. How many ml of each vinegar do you need to m
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The problem: one is 5% acid and one is 6.5% acid . You want to make 200ml of vinegar,with 6% acid. How many ml of each vinegar do you need to m
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Question 679506: Solving Linear Systems of equations
The problem: one is 5% acid and one is 6.5% acid . You want to make 200ml of vinegar,with 6% acid. How many ml of each vinegar do you need to mix together?
My question: With percents I get confused, and I am not sure which variables you would use to set this up into two equations? Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! Let quantity of 5% acid required be x
Let quantity of 6.5% acid required be y
Mixture = 6%
Equate percentages
5%x+6.5%y=6%*200
multiply by 100
5x+6.5y=1200................(1)
Quantity
x+y=200.....................(2)
multiply equation (2) by -5
we get -5x-5y=-1000
Add this equation to (1)
we get
1.5y=200
y= 200/1.5
y=133.3 ml
x= balance amount= 66.67ml
