Question 4 Consider the following three systems of linear equations. \[ \begin{array}{c} \text { System A } \\ \left\{\begin{array} { c } { \text { System B } } \\ { - 2 x + 3 y = - 7 [ \mathrm { A } 1 ] } \\ { - 7 x + 4 y = 8 } \end{array} \left[\begin{array} { c } { \text { [A2] } } \end{array} \left\{\begin{array} { r } { - 2 x + 3 y = - 7 [ \mathrm { B } 1 ] } \\ { 2 1 x - 1 2 y = - 2 4 [ \mathrm { B } 2 ] } \end{array} \left\{\left\{\begin{array}{c} -2 x+3 y=-7[\mathrm{C} 1] \\ 13 x=-52 \end{array}[\mathrm{C} 2]\right.\right.\right.\right.\right. \end{array} \] 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 \( [\mathrm{A} 1] \rightarrow \) Equation \( [ \) B1] \( -3 \times \) Equation [A2] \( \rightarrow \) Equation [B2] \( \square \) \( \times \) Equation \( [\mathrm{A} 1]+ \) Equation \( [\mathrm{A} 2] \rightarrow \) Equation \( [\mathrm{B} 2] \) \( \square \) \( \times \) Equation \( [A 2]+ \) Equation \( [A 1] \rightarrow \) Equation \( [B 1] \) (b) How do we transform System B into System C? \( \times \) Equation [B1] \( \rightarrow \) Equation [C1] \( \square \) \( \times \) Equation [B2] \( \rightarrow \) Equation [C2] \( \square \) \( \times \) Equation \( [\mathrm{B} 1]+ \) Equation \( [\mathrm{B} 2] \rightarrow \) Equation \( [\mathrm{C} 2] \) \( \square \) \( \times \) Equation \( [\mathrm{B} 2]+ \) Equation \( [\mathrm{B} 1] \rightarrow \) Equation \( [\mathrm{C} 1] \)

We are given three systems:  • System A:   [ A1 ] –2x + 3y = –7   [ A2 ] –7x + 4y = 8  • System B:   [ B1 ] –2x + 3y = –7         (same as A1)   [ B2 ] 21x – 12y = –24  • System C:   [ C1 ] –2x + 3y = –7         (same as B1)   [ C2 ] 13x = –52 We now explain the two parts. ────────────────────────────── Part (a). Transform System A into System B. Look at the two equations of System A and B: • Equation [A1] is exactly the same as [B1] so no change is needed (i.e. multiplying by 1). • Equation [A2] is –7x + 4y = 8; notice that if you multiply [A2] by –3 you get   (–3)(–7x) = 21x, (–3)(4y) = –12y, (–3)(8) = –24, which is exactly Equation [B2]. Thus, one acceptable answer is:   1× Equation [A1] → Equation [B1]  and  (–3)× Equation [A2] → Equation [B2]. ────────────────────────────── Part (b). Transform System B into System C. Comparing the equations: • Equation [B1] is –2x + 3y = –7, and this is identical to Equation [C1]. (That is, 1× Equation [B1] gives Equation [C1].) • Equation [C2] is 13x = –52, and it must be obtained from Equation [B2] together with [B1] in order to “eliminate” y. Notice that:  – Equation [B2] is 21x – 12y = –24.  – If we multiply Equation [B1] (–2x + 3y = –7) by 4 we get: –8x + 12y = –28. Now, adding these results, we have:   (21x – 12y) + (–8x + 12y) = (13x)  and  (–24) + (–28) = –52. Thus, 4× Equation [B1] + Equation [B2] gives Equation [C2]. An acceptable answer is:   1× Equation [B1] → Equation [C1]  and  4× Equation [B1] + Equation [B2] → Equation [C2]. ────────────────────────────── Summary of Answers (a) To go from System A to System B:   • Multiply Equation [A1] by 1 (that is, leave it unchanged) to get Equation [B1].   • Multiply Equation [A2] by –3 to get Equation [B2]. (b) To go from System B to System C:   • Multiply Equation [B1] by 1 (leave it unchanged) to get Equation [C1].   • Multiply Equation [B1] by 4 and add the result to Equation [B2] to obtain Equation [C2]. Any answer equivalent to the one above is correct.