Consider the following three systems of linear equations. $$ \begin{array}{c} \text { System A } \ \left\{\begin{array} { c } { - 4 x + 3 y = 2 3 [ \mathrm { A } 1 ] } \ { 2 x - 7 y = 5 } \end{array} \left\{\begin{array}{c} -11 y=33[\mathrm{~B} 1] \ 2 x-7 y=5[\mathrm{~B} 2] \end{array}\left\{\begin{array}{c} y=-3 \ 2 x-7 y=5[\mathrm{C} 1] \end{array}\right]\right.\right. \end{array} $$ 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 $$ [\mathrm{B} 1] $$ $$ \square $$ $$ \times $$ Equation $$ [\mathrm{A} 2] \rightarrow $$ Equation [B2] $$ \square $$ $$ \times $$ Equation $$ [\mathrm{A} 1]+ $$ Equation $$ [\mathrm{A} 2] \rightarrow $$ Equation $$ [\mathrm{B} 2] $$ $$ \square $$ $$ \times $$ 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 $$ [\mathrm{B} 1] \rightarrow $$ Equation [C1] $$ \square $$ $$ \times $$ Equation [B2] $$ \rightarrow $$ Equation [C2] $$ ] \times $$ Equation $$ [\mathrm{B} 1]+ $$ Equation $$ [\mathrm{B} 2] \rightarrow $$ Equation [C2] $$ \square $$ $$ \times $$ Equation $$ [B 2]+ $$ Equation $$ [B 1] \rightarrow $$ Equation $$ [C 1] $$

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We wish to obtain System B from System A and then System C from System B by doing simple algebraic manipulations. (Note that all the “×” signs in the options indicate you are multiplying one equation by a constant and possibly then adding another equation.)

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Step 1. Transforming System A into System B

System A is:
  [A1]: –4x + 3y = 23 
  [A2]: 2x – 7y = 5

System B is:
  [B1]: –11y = 33 
  [B2]: 2x – 7y = 5

Notice that B2 is the same as A2 (so it is unchanged) and that B1 is an equation with just y. In fact, you can “eliminate” the x‐term from A1 and A2 as follows. Multiply A2 by 2:
  2 × (2x – 7y) = 4x – 14y  with right side 2×5 = 10.

Now add this to A1:
  [–4x + 3y] + [4x – 14y] = (–4x + 4x) + (3y – 14y) = –11y,
  and 23 + 10 = 33.

Thus we have obtained –11y = 33, which is exactly B1.

So the transformation for part (a) is:
  • 2 × Equation [A2] + Equation [A1] → Equation [B1]  
  • 1 × Equation [A2] → Equation [B2]

That is, you use “2 × [A2] + [A1] → [B1]” and leave A2 as it is for [B2].

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Step 2. Transforming System B into System C

System B is:
  [B1]: –11y = 33 
  [B2]: 2x – 7y = 5

System C is:
  [C1]: y = –3 
  [C2]: 2x – 7y = 5

Clearly, C2 is identical to B2. For C1, we solve B1 for y by multiplying by –1/11:
  (–1/11)×(–11y) = y  and (–1/11)×33 = –3.

Thus the transformation for part (b) is:
  • (–1/11) × Equation [B1] → Equation [C1]
  • 1 × Equation [B2] → Equation [C2]

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Final Answers

(a) To go from System A to System B:
  – Multiply Equation [A2] by 2 and add it to Equation [A1] to get Equation [B1];
  – Equation [A2] is unchanged (i.e. multiplied by 1) to give Equation [B2].

(b) To go from System B to System C:
  – Multiply Equation [B1] by –1/11 to get Equation [C1];
  – Equation [B2] is unchanged (i.e. multiplied by 1) to yield Equation [C2].
(a) To transform System A into System B: - Multiply Equation [A2] by 2 and add it to Equation [A1] to get Equation [B1]. - Equation [A2] remains unchanged to become Equation [B2]. (b) To transform System B into System C: - Multiply Equation [B1] by –1/11 to get Equation [C1]. - Equation [B2] remains unchanged to become Equation [C2].

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