Çözüldü Problem Solution by Using Multivariable Equation System

A solved example from San Diego State University:

A man has twenty coins consisting of dimes and quarters. If the dimes were quarters and the quarters were dimes, he would have ninety cents more than he has now. How many dimes and quarters does he have?

Let x be the number of dimes, y be the number of quarters, and z be the amount of money that the man has.
It follows that x + y = 20, 0.10x + 0.25y = z, and 0.25x + 0.10y = z + 0.9.
The first equation gives y = 20 − x, which we substitute into the other equations giving
0.10x + 0.25(20 − x) = z and 0.25x + 0.10(20 − x) = z + 0.9.
It follows that z + 0.15x = 5 and z − 0.15x = 1.1.
Subtracting the second equation from the first, gives 0.3x = 3.9 or x = 13.
It follows that y = 7 and z = $3.05.

https://jmahaffy.sdsu.edu/courses/s00/math121/introduction/selftest.pdf (Son soru)
https://jmahaffy.sdsu.edu/courses/s00/math121/introduction/algsolnlong.pdf (Çözümler)

Not:
The number of dimes: x = 13
The number of quarters: y = 7
Total money the man has: x = 10x + 25y = 10·13 + 25·7 = 305 cents = 3 dollars and 5 cents