x = number of programmable calculators
y = number of scientific calculators
z = number of household type calculators
Total cost is $2000
→ Equation (1)
Total number of units is 20
→ Equation (2)
No other equation can be made from the problem. Although the number of equations is less than the number of unknowns (indeterminate), variables x, y, and z can only hold positive whole numbers, thus, we can solve the problem.
Subtract Equation (2) from Equation (1)
By trial and error:
Try x = 1, y = 7.5 → not applicable
Try x = 2, y = 5 → okay!
Try x = 3, y = 2.5 → not applicable
Try x = 4, y = 0 → not acceptable
Substitute x = 2 and y = 5 to Equation (2)
Thus, the man bought 2 units of programmable calculators, 5 units of scientific calculators, and 13 units of household type calculators.