Im confused by this

Question

In a collection of dimes and quarters, there are 48 coins. The total value of the coins is $7.50. How many coins of each type are there?

Answer 1

You have two unknowns, # of dimes and quarters, so you need two equations to solve this. So, let's let x=# dimes and y=# of quarters.

Equation 1:  x+y=48

Equation 2:  10x+25y=750

You can use either the substitution method or the elimination method to solve this. Let's use the substitution method:

Solve for x in the first equation: x+y=48

                                               x=48-y

Use this value for x in the second equation and solve for y:

                                               10x+25y=750

                                               10(48-y)+25y=750

                                               480-10y+25y=750

                                               -10y+25y=750-480

                                               15y=270

                                                  y=18

Put this value of y in Equation 1 to solve for x:

                                               x+y=48

                                               x+18=48

                                               x=48-18

                                               x=30

Check:  x+y=48                       10x+25y=750

            30+18=48                   10(30)+25(18)=750

            48=48                         300+450=750

            Correct!                       750=750  Correct

Therefore, there are 30 dimes and 18 quarters.