First, let's call the number of nickels we have #n# and the number of quarters #q#.
Now we can write two equations which we can use to solve the problem through substitution:
First we know there are 22 coins in total, therefore:
#n + q = 22#
And we know their value is $3.70 so we can write:
#0.05n + 0.25q = 3.70#
Step 1) solve the first equation for #n#
#n + q = 22#
#n + q - color(red)(q) = 22 - color(red)(q)#
#n + 0 = 22 - color(red)(q)#
#n = 22 - q#
Step 2) Substitute #22 - q# for #n# in the second equation and solve for #q#.
#0.05(22 - q) + 0.25q = 3.70#
#(0.05 xx22) - (0.05 xx q) + 0.25q = 3.70#
#1.1 - 0.05q + 0.25q = 3.70#
#1.1 + 0.25q - 0.05q = 3.70#
#1.1 + (0.25 - 0.05)q = 3.70#
#1.1 + 0.20q = 3.70#
#1.1 - color(red)(1.1) + 0.20q = 3.70 - color(red)(1.1)#
#0 + 0.20q = 2.60#
#0.20q = 2.60#
#(0.20q)/color(red)(0.20) = 2.60/color(red)(0.20)#
#(color(red)(cancel(color(black)(0.20)))q)/cancel(color(red)(0.20)) = 13#
#q = 13#
Step 3) Substitute #13# for #q# in the solution to the first equation in Step 1.
#n = 22 - 13#
#n = 9#
Solution:
There are 9 nickels and 13 quarters
