A pharmacist is mixing a 3%saline sol.& an 8%saline sol.to get 2 litres of a 6% sal.sol. how much of each sol.

a math question .please help.

Answer:
Use X for the amount of 3% sol, and Y for the amount of 8% sol. Both in litres

3X + 8Y = 6 * 2

Remember that we know that X + Y = 2, so we can substitute for one of them (I chose X).

X = 2 - Y

Use that in the original equation,

3 * (2 - Y) +8Y = 12

Solve for Y,

6 + 5Y = 12

5Y = 12 - 6 = 6

Y = 6/5

Going back to X + Y = 2,

X + 6/5 = 2

X = 2 - 6/5 = 4/5

Check it out in your problem

3 * (4/5) + 8 * (6/5) = 12/5 + 48/5 = 12

Multiply through by 5

12 + 48 = 12 * 5 = 60

That checks out, so X (3% solution) = 4/5 litres, and Y (8% solution) = 6/5 litres
lets x be the volume of 3%, so the volume of 8% is (2-x)

and you write if you use percentages

3x+ (2-x) 8 = 6*2=12
3x-8x+16 =12
-5x=-4 x=0.8l and 1-x =1.2

you use 0.8litres of 3% and 1.2 litres of 8%
(8 - 6) / (8 - 3) = 2/5 of the 3% solution is in the final concentration.
Times 2 liters needed = 4/5 liter of 3%, and 1 1/5 liter of 8%
1200ml of 8% + 800ml of 3% = 2000ml of 6%

1200(.08) + 800(.03) = 2000(.06)

96 + 24 = 120

120 = 120
Consider the following set-up:

you are using 'x' ml of the 3% and '2000 minus x' ml of the 8%

together they make 2000 ml of 6%

That's all the help you deserve

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