Number of crayons have Peg: x=?
Number of crayons have Pat: y=?
Peg and Pat are sharing 64 crayons:
(1) x+y=64
Pat has 10 more crayons than Peg:
(2) y=x+10
We have a system of 2 equations and 2 unkowns. Using the method of substitution:
Replacing (2) y=x+10 in equation (1):
(1) x+y=64
x+(x+10)=64
x+x+10=64
2x+10=64
Solving for x:
2x+10-10=64-10
2x=54
2x/2=54/2
x=27
Replacing x=10 in equation (2)
(2) y=x+10
y=27+10
y=37
Answer:
Peg has 27 crayons
Pat has 37 crayons