4a) Let Fred’s age now be represented by F and Bill’s age now be represented by B.
F - 25 = B (1)
F + 5 = 2(B+5) (2)
From (1), F = B + 25, substitute this in (2).
B+25+5 = 2(B+5)
B + 30 = 2B + 10
2B - B = 30 - 10
B = 20 (3)
Substitue (3) in (1),
F - 25 = 20
F = 20 + 25
= 45
4b) Let the number of 5c be represented as x and 10c be represented as y.
x+y = 50 (1)
0.05x + 0.10y = 4.20 (2)
From (1), x = 50-y. Substitute this into (2).
0.05(50-y) + 0.10y = 4.20
2.50-0.05y+0.10y = 4.20
0.05y = 1.70
y = 1.70/0.05
y = 34 (3)
Substitute (3) into (1)
x = 50-34
x = 16
He has 16 5c and 34 10c.