.A tank contains seven liters of 10% alcohol solution. How many liters of 6% alcohol solution should be added to have 8% alcohol solution?

The amount of 6% alcohol solution should be added to 7 liters of 10% alcohol solution to have 8% alcohol solution is 7 liters.

An equation is formed when two equal expressions are equated together with the help of an equal sign '='.

Let the amount of 6% mixed in the 10% alcohol be x.

We know that the total solution is 7 liter of 10% alcohol, and x liters of 6% alcohol. And we need to make 8% alcohol solution, therefore, the total alcohol in the mixed solution can be written as,

Hence, the amount of 6% alcohol solution should be added to 7 liters of 10% alcohol solution to have 8% alcohol solution is 7 liters.

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7 liters of 6% alcohol solution.

Step-by-step explanation:

We have 7 liters of 10% alcohol solution, and in this the total amount of alcohol is

A = (10%/100%)*7 L = 0.1*7L = 0.7L

Now, suppose that we add N liters of 6% alcohol solution, then we have a total of:

(N + 7) Liters

And the total amount of alcohol would be:

(6%/100%)*N + 0.7L

0.06*N + 0.7L

The quotient between the total volume of alcohol and the total volume of liquid needs to be 0.08 (because we want a percentage of 8% of alcohol, and the decimal form of 8% is 0.08)

Then:

(0.06*N + 0.7L)/(N + 7L) = 0.08

We want to sove this for N

(0.06*N + 0.7L) = 0.08*(N + 7L)

0.06*N + 0.7L = 0.08*N + 0.56L

0.7L - 0.56L = 0.08*N - 0.06*N

0.14 L = 0.02*N

0.14L/0.02 = N = 7L

Then we need to add 7 liters of 6% alcohol solution.

the correct answer is 828.07

y=3n + 1

step-by-step explanation:

this is wrong because it says the second number is one less and i added one instead