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You have to think:
What we want?
50% antifreeze solution.
In other words
“to obtain final solution that contains an equal volume of antifreeze and water”
We can translate this to an equation:
Vf,anti =Vf,water
We have two different initial solutions that will be mixed:
8 litters with 70 % of antifeeze
and
X liters with 40 % of antifreeze. (X is the variable we want to calculate)
translating to an equations we have:
for the 8 liter:
Vi,ant i= 8 x 0.7 = 5.6 liters of antifreeze and
Vi,water = 8 x 0.3 =2.4 liters of Water
for the 40%
Vadd, anti = X x 0.4 liters of antifreeze
Vadd, water = X x 0.6 liters of water
The final total volume of water and antifreeze that we have afer mixter is
Vf,ant i= Vadd,anti + Vi,anti
Vf,water = Vadd,water + Vi,water
sbstutuing the variables
Vf,anti = 8 x 0.7 + X x 0.4
Vf,water= 8 x 0.3 + X x0.6
So now we determinated the Variables os the first equation:
subtituing the variable in the first equation we have:
8 x 0.7 + X x 0.4 = 8 x 0.3 + X x 0.6
Isolate the X variable.
Xx0.4 - Xx0.6=8*0.3-8x0.7
X(0.4-0.6)= 2.4-5.6
Multiplying the equation by -1
0.2X=3.2
X=3.2/0.2
X=16 liters
Are you asking for 4 Liters of pure (100%) antifreeze? Here is a lesson which fits your (this) form:
Two-Part Mixture: One material Quantity Unknown
If you find a result, check the result in your original equation to find if it works or not. To start, Let x = volume of water to add.