**A = (15 - 50k)/(0.15 - k) and B = (-7.5)/(0.15 - k)**

how much 15% antifreeze should be mixed to get 50 gallons of 30% antifreeze

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There is one chemical on hand that contains 15% antifreeze. Call this chemical A.

Let k be the percent of anti-freeze in the other chemical that is NOT given here. Call this chemical B.

Note that k is not 15%, for otherwise it would be the same as chemical A.

Let A and B represent the amounts of each chemical needed.

A+B = 50 ---> B = 50-A

The amount of anti-freeze is 0.15A + kB = 0.3*50 = 15

0.15A + k(50-A) = 15

0.15A + 50k - Ak = 15

0.15A - Ak = 15 - 50k

(0.15-k)A = 15 - 50k

A = (15 - 50k)/(0.15 - k)

B = 50 - A = 50 - (15 - 50k)/(0.15-k)

= [50(0.15 - k) - (15 - 50k) ]/(0.15 - k)

= [ 7.5 - 50k - 15 + 50k]/(0.15 - k)

= (-7.5)/(0.15 - k)

So in terms of k,

Now we test it.

Suppose that chemical B contains k=50%=0.5 anti-freeze

Then A = (15 - 50*0.5)/(0.15 - 0.5) = (15 - 25)/(0.15-0.5) = (-10)/(-0.35) = 28 and 4/7

B = (-7.5)/(0.15 - 0.5) = (-7.5)/(-0.35) = 21 and 3/7

Note that A+B=50 and

(0.15)(28 and 4/7) + (0.5)(21 and 3/7) = 4 and 2/7 + 10 and 5/7 = 15 = 50*0.3 = 30% of 50

So yes, the formulas work.

So WHEN you find out what percent of anti-freeze, k, is in the second chemical (chemical B)

you plug it into the formulas I have derived for in bold above to get your answer.

We have given this problem it's due diligence, and it is the best we can do for you.

Good luck.

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