[b]Assignment.[/b] A company has two filter types in storage, but the unit prices for the filters are missing. According to the logs, the company bought 50 type A filters and 100 type B filters on January, with a total cost of 450 euros. On February, the company bought 30 type A filters and 40 type B filters, with a total cost of 200 €. Use this information to recover the unit prices for each filter.[br][br][b]Solution.[/b] Let us denote the unit prices [math]A[/math] (for filter A) and [math]B[/math] (for filter B). If we buy 50 type A filters, their price is [math]50A[/math], and if we buy 100 type B filters, their price is [math]100B[/math]. Therefore, the total costs for January can be expressed with the equation[br][br][math] \large 50A + 100B = 450 , [/math][br][br]and similarly for February. We obtain a pair of equations:[br][br][math] \large \left\{ \begin{array}{rcl}[br]50A + 100B & = & 450 \\[br]30A + 40B & = & 200 \\[br]\end{array} \right. [/math][br][br]Let us multiple the equations so that the coefficients of [math]A[/math] are opposite numbers:[br][math] \large \left\{ \begin{array}{rcll}[br]50A + 100B & = & 450 & | \cdot 30 \\[br]30A + 40B & = & 200 & | \cdot (-50) \\[br]\end{array} \right. [/math][br][br][math] \large \left\{ \begin{array}{rcl}[br]1500A + 3000B & = & 13500 \\[br]-1500A - 2000B & = & -10000 \\[br]\end{array} \right. [/math][br][br]Adding rows 1 and 2 yields[br][br][math] \large \begin{array}{rcll}[br]1000B & = & 3500 & | : 1000 \\[br]B & = & \frac{3500}{1000} = 3,5[br]\end{array} [/math][br][br]Substituting this on the 1st row yields[br][br][math] \large \begin{array}{rcll}[br]50A + 100 \cdot 3,5 & = & 450 \\[br]50A + 350 & = & 450 \\[br]50A & = & 100 & | : 50 \\[br]A & = & \frac{100}{50} = 2 [br]\end{array} [/math][br][br]The unit prices are 2 euros for filter A, and 3,5 euros for filter B.