**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: **

** **

### V_{f,anti} =V_{f,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:*

**V**_{i,ant i}= 8 x 0.7 = 5.6 liters of antifreeze and

**V**_{i,water }= 8 x 0.3 =2.4 liters of Water

*for the 40%*

**V**_{add, anti }= X x 0.4 liters of antifreeze

**V**_{add, water = }X x 0.6 liters of water

*The final total volume of water and antifreeze that we have afer mixter is *

** V**_{f,ant i}= V_{add,anti} + V_{i,anti}

** V**_{f,water} = V_{add,water} + V_{i}_{,water}

*sbstutuing the variables*

**V**_{f,anti} = 8 x 0.7 + X x 0.4

**V**_{f,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

## Comments

THANK YOU VERY MUCH!