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 $$ [\mathrm{C} 1] $$

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UpStudy AI · Accepted Answer

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.

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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].

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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].

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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.
(a) To transform System A into System B: - Multiply Equation [A1] by 1 to get Equation [B1]. - Multiply Equation [A2] by –3 to get Equation [B2]. (b) To transform System B into System C: - Multiply Equation [B1] by 1 to get Equation [C1]. - Multiply Equation [B1] by 4 and add it to Equation [B2] to get Equation [C2].

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