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how many gallons of antifreeze do I need to add to 196 gallons of water to get 
20% antifreeze mixture ?
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First, pay attention on how I edited your post to make it correct.
It can be solved mentally.
If the mixture contains 20% of pure antifreeze, it means that the mixture contains 0.8 = 80% water of the total volume.
From it, you can calculate the total volume in one line: Total volume = 
= 245 gallons.
In the second line, you calculate how much of the pure antifreeze should be added: 245 - 196 = 49 gallons.
Answer. 49 gallons of the pure antifreeze should be added.
Algebra solution is as follows:
= 0.2 <<<---=== it is the "concentration" equation, saying that
The ratio of the antifreeze volume to the total volume is 0.2, the assigned concentration
x = 0.2*(196 + x)
x = 0.2*196 + 0.2x
x - 0.2x = 0.2*196
0.8x = 0.2*196 ====> x = 
= 49 gallons.
You get the same answer.
-------------------
It is a standard and typical mixture problem.
For introductory lessons covering various types of mixture word problems see
- Mixture problems
- More Mixture problems
- Solving typical word problems on mixtures for solutions
- Typical word problems on mixtures from the archive
- Word problems on mixtures for antifreeze solutions
in this site.
You will find there ALL TYPICAL mixture problems with different methods of solutions,
explained at different levels of detalization, from very detailed to very short.
Read them and become an expert in solution mixture word problems.
Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this textbook in the section "Word problems" under the topic "Mixture problems".
Save the link to this online textbook together with its description
Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson
to your archive and use it when it is needed.
