how many of each kind were there?

Question

A jar contained $55 in nickels and dimes. If there were 800 coins in the jar

Answer 1

X=nickels and dimes =Y

Eqt.1) X+Y=800

Eqt.2) .10X+.05Y=500

Find nickels first, solve Eqt.1 for "y" and plug into Eqt. 2

Y=800-x

.10X+.05(800-X)=55

.10X+40-.04X=55

.05X+40=55

X= (55-40)/.05

X=300 nickels

Find Dimes

Y=800-X

y=800-300

y=500 Dimes