Let

*x* = number of programmable calculators

*y* = number of scientific calculators

*z* = number of household type calculators

Total cost is \$2000

$300x + 150y + 50z = 2000$

$6x + 3y + z = 40$ → Equation (1)

Total number of units is 20

$x + y + z = 20$ → Equation (2)

Note:

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)

$5x + 2y = 20$

$y = \dfrac{20 - 5x}{2}$

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)

$2 + 5 + z = 20$

$z = 13$ *answer*

Thus, the man bought 2 units of programmable calculators, 5 units of scientific calculators, and 13 units of household type calculators.