Consider the following three systems of linear equations. \[ \begin{array}{c} \text { System A } \\ \left\{\begin{array} { c c } { - 8 x + 5 y = 4 [ \mathrm { A } 1 ] } \\ { - 7 x + 2 y = 1 3 [ \mathrm { A } 2 ] } \end{array} \left\{\begin{array} { c } { 6 x + y = - 2 2 [ \mathrm { B } 1 ] } \\ { - 7 x + 2 y = 1 3 [ \mathrm { B } 2 ] } \end{array} \left\{\left\{\begin{array}{c} 6 x+y=-22[\mathrm{C} 1] \\ -19 x=57 \end{array}[\mathrm{C} 2]\right.\right.\right.\right. \end{array} \] Answer the questions below. For each, choose the transformation and then fill in the blank with the correct number. The arrow ( \( \rightarrow \) ) means the expression on the left becomes the expression on the right. (a) How do we transform System A into System B? \( \square \) \( \times \) Equation [A1] \( \rightarrow \) Equation [B1] \( \square \) \( \times \) Equation [A2] \( \rightarrow \) Equation [B2] \( \square \) \( \times \) Equation \( [\mathrm{A} 1]+ \) Equation \( [\mathrm{A} 2] \rightarrow \) Equation [B2] \( \square \) \( \times \) Equation \( [\mathrm{A} 2]+ \) Equation \( [\mathrm{A} 1] \rightarrow \) Equation \( [\mathrm{B} 1] \) (b) How do we transform System B into System C? \( \square \) \( \times \) Equation \( [B 1] \rightarrow \) Equation [C1] \( ] \times \) Equation \( [B 2] \rightarrow \) Equation [C2] \( \square \) \( \times \) Equation \( [B 1]+ \) Equation \( [B 2] \rightarrow \) Equation [C2] \( \square \) \( \times \) Equation \( [B 2]+ \) Equation \( [B 1] \rightarrow \) Equation \( [C 1] \)

We are given three systems of equations whose relationships are expressed by transforming one system into the next. One acceptable answer (with explanation) is as follows. ───────────────────────────── Step (a): Transforming System A into System B System A:   [A1] –8x + 5y = 4   [A2] –7x + 2y = 13 System B:   [B1] 6x + y = –22   [B2] –7x + 2y = 13 Notice that Equation [B2] is exactly the same as Equation [A2]. Hence, to go from A to B we do nothing to equation [A2] (or, equivalently, multiply it by 1). The only change is to obtain [B1] from the given equations. We want to “convert” [A1] (–8x + 5y = 4) into 6x + y = –22. In fact, one may verify that if we take   [A1] + (–2)×[A2], then the left–sides combine as follows:   (–8x + 5y) + (–2)(–7x + 2y) = –8x + 5y + 14x – 4y = 6x + y. And the constants give:   4 + (–2)·13 = 4 – 26 = –22. Thus the transformation is:   (–2)× Equation [A2] + Equation [A1] → Equation [B1] and  1× Equation [A2] → Equation [B2]. One may fill the blanks as follows:   □ × Equation [A2] + Equation [A1] → Equation [B1] with the blank filled by –2.   □ × Equation [A2] → Equation [B2] (with the blank 1). ───────────────────────────── Step (b): Transforming System B into System C System B:   [B1] 6x + y = –22   [B2] –7x + 2y = 13 System C:   [C1] 6x + y = –22   [C2] –19x = 57 Notice that Equation [C1] is exactly the same as [B1] (so it is “left alone”). To produce [C2] we must eliminate y from the equations. One method is to combine B1 and B2. In fact, if we take –2 times [B1] and add [B2] we have:   –2×(6x + y) + (–7x + 2y) = –12x – 2y – 7x + 2y = –19x. On the right–hand side, we have:   –2×(–22) + 13 = 44 + 13 = 57. Thus the transformation is:   (–2)× Equation [B1] + Equation [B2] → Equation [C2] and  1× Equation [B1] → Equation [C1]. One may fill the blanks as follows:   □ × Equation [B1] → Equation [C1] (with the blank 1)   □ × Equation [B1] + Equation [B2] → Equation [C2] (with the blank –2). ───────────────────────────── Summary of Answers (a) To transform System A into System B:  • Equation [B2] is obtained by multiplying [A2] by 1:   1 × [A2] → [B2].  • Equation [B1] is obtained by multiplying [A2] by –2 and then adding [A1]:   (–2) × [A2] + [A1] → [B1]. (b) To transform System B into System C:  • Equation [C1] is obtained by multiplying [B1] by 1:   1 × [B1] → [C1].  • Equation [C2] is obtained by multiplying [B1] by –2 and then adding [B2]:   (–2) × [B1] + [B2] → [C2]. Any answer equivalent to the one above is correct.