.
If the number of dimes is x, then the number of quarters is (x-2).
The total "money" equation is
10*x + 25*(x-2) = 195 cents, or
10x +25x - 50 = 195
35x = 195 + 50 = 245
x = 
= 7.
Answer. 7 dimes and 7-2 = 5 quarters.
CHECK. 7*10 + 5*25 = 195 cents. ! Precisely correct !
Solved.
You can solve it mentally, too.
Simply put (mentally) 2 dimes aside, for a minute.
Then your updated collection contains equal number of dimes and quarters and is worth 195 - 2*10 = 175 cents.
You can group the coins in the reduced collection in pairs containing 1 quarter and 1 dime each.
Each pair is worth 10+25 = 35 cents and the number of pairs is 
= 5.
You get the same answer: 5 quarters and 5+2 = 7 dimes.
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For coin problems and their detailed solutions see the lessons in this site:
- Coin problems
- More Coin problems
- Solving coin problems without using equations
- Kevin and Randy Muise have a jar containing coins
- Typical coin problems from the archive
- Three methods for solving standard (typical) coin word problems
- More complicated coin problems
- Solving coin problems mentally by grouping without using equations
- Non-typical coin problems
- Santa Claus helps solving coin problem
You will find there the lessons for all levels - from introductory to advanced,
and for all methods used - from one equation to two equations and even without equations.
Read them attentively and become an expert in this field.
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 online textbook under the topic "Coin 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.
