• There are four unknowns:
  Let p = the number of pennies
  Let n = the number of nickels
  Let d = the number of dimes
  Let q = the number of quarters
• With four unknowns we will need a system of four equations to solve the problem.
• Since there are 513 coins altogether:
  p + n + d + q = 513
• Since the number of pennies is twice the sum of nickels, dimes and quarters:
  p = 2*(n + d + q)
  or
  p = 2n + 2d + 2q
• Since the weight of everything is 1779 grams and the weight of the jar is 301, the weight of all the coins is 1779 - 301 = 1478 grams.
• Since each penny weighs 2.5g, "p" pennies would weigh 2.5p grams. Since each nickel weighs 5g, "n" nickels would weigh 5n grams. Since each dime weighs 2.3g, "d" dimes would weigh 2.3d grams. Since each quarter weighs 5.8g, "q" pennies would weigh 5.8q grams. And finally, since all the coins together weigh 1478 grams:
  2.5p + 5n + 2.3d + 5.8q = 1478
  If you don't want to work with decimals, you can multiply both sides by 10:
  25p + 50n + 23d + 58q = 14780
  
• The final equation will deal the the values of these coins. To make things easier (IMHO), I am going to use values expressed in terms of cents instead of dollars. So instead of the total value being $17.12 I will use 1712 cents. Each penny is worth 1 cent so p pennies will be worth p cents. Each nickel is worth 5 cents so n nickels will be worth 5n cents. Each dime is worth 10 cents so d dimes will be worth 10d cents. Each quarter is worth 25 cents so q quarters will be worth 25q cents. So
  p + 5n + 10d + 25q = 1712
  If you'd rather work with dollars, this equation would be:
  0.01p + 0.05n + 0.10d + 0.25q = 17.12

p + n + d + q = 513
p = 2*(n + d + q)               or   p = 2n + 2d + 2q
2.5p + 5n + 2.3d + 5.8q = 1478  or   25p + 50n + 23d + 58q = 14780
p + 5n + 10d + 25q = 1712       or   0.01p + 0.05n + 0.10d + 0.25q = 17.12

• Substitution
• Linear combination (aka elimination, aka addition)
• Several matrix-based methods (Gaussian elimination, inverse matrices, etc.)
• Determinants/Cramer's Rule