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Consider the following three systems of linear equations:

System A:
[tex]\[
\begin{cases}
-3x + 3y = 6 \quad [\mathrm{A}1] \\
7x - 4y = -2 \quad [\mathrm{A}2]
\end{cases}
\][/tex]

System B:
[tex]\[
\begin{cases}
-3x + 3y = 6 \quad [\mathrm{B}1] \\
x + 2y = 10 \quad [\mathrm{B}2]
\end{cases}
\][/tex]

System C:
[tex]\[
\begin{cases}
9y = 36 \quad [\mathrm{C}1] \\
x + 2y = 10 \quad [\mathrm{C}2]
\end{cases}
\][/tex]

Answer the questions below. For each, choose the transformation and then fill in the blank with the correct number. The arrow [tex]$(\rightarrow)$[/tex] means the expression on the left becomes the expression on the right.

(a) How do we transform System A into System B?

1. [tex]\(\square \times\)[/tex] Equation [tex]$[\mathrm{A}1] \rightarrow$[/tex] Equation [tex]$[\mathrm{B}1]$[/tex]
2. [tex]\(\square \times\)[/tex] Equation [tex]$[\mathrm{A}2] \rightarrow$[/tex] Equation [tex]$[\mathrm{B}2]$[/tex]
3. [tex]\(\square \times\)[/tex] Equation [tex]$[\mathrm{A}1] +$[/tex] Equation [tex]$[\mathrm{A}2] \rightarrow$[/tex] Equation [tex]$[\mathrm{B}2]$[/tex]
4. [tex]\(\square \times\)[/tex] Equation [tex]$[\mathrm{A}2] +$[/tex] Equation [tex]$[\mathrm{A}1] \rightarrow$[/tex] Equation [tex]$[\mathrm{B}1]$[/tex]

(b) How do we transform System B into System C?

1. [tex]\(\square \times\)[/tex] Equation [tex]$[\mathrm{B}1] \rightarrow$[/tex] Equation [tex]$[\mathrm{C}1]$[/tex]
2. [tex]\(\square \times\)[/tex] Equation [tex]$[\mathrm{B}2] \rightarrow$[/tex] Equation [tex]$[\mathrm{C}2]$[/tex]
3. [tex]\(\square \times\)[/tex] Equation [tex]$[\mathrm{B}1] +$[/tex] Equation [tex]$[\mathrm{B}2] \rightarrow$[/tex] Equation [tex]$[\mathrm{C}2]$[/tex]
4. [tex]\(\square \times\)[/tex] Equation [tex]$[\mathrm{B}2] +$[/tex] Equation [tex]$[\mathrm{B}1] \rightarrow$[/tex] Equation [tex]$[\mathrm{C}1]$[/tex]

Sagot :

GinnyAnsweridn
Let's carefully break down the transformations needed to go from System A to System B and from System B to System C.

### Transformation from System A to System B:

System A:
[tex]\[ \begin{cases} -3x + 3y = 6 \quad [\text{A1}] \\ 7x - 4y = -2 \quad [\text{A2}] \end{cases} \][/tex]

System B:
[tex]\[ \begin{cases} -3x + 3y = 6 \quad [\text{B1}] \\ x + 2y = 10 \quad [\text{B2}] \end{cases} \][/tex]

Given this, we need to see how we transform each equation from System A to System B.

- Equation [tex]\([\text{A1}]\)[/tex] is already equal to [tex]\([\text{B1}]\)[/tex], so there is no change needed. This means we multiply [tex]\([\text{A1}]\)[/tex] by 1.
[tex]\[ \times \text{Equation} [\text{A1}] \rightarrow \text{Equation} [\text{B1}] \][/tex]

- To transform [tex]\([\text{A2}]\)[/tex] to [tex]\([\text{B2}]\)[/tex], we need a completely different equation, since [tex]\([\text{B2}]\)[/tex] is [tex]\( x + 2y = 10\)[/tex].

Given this, we can see that we have taken the equation in [tex]\([\text{B2}]\)[/tex] directly. For the sake of simplicity, assume that [tex]\([\text{A2}]\)[/tex] needs a factor of 1 multiplier to be used directly in [tex]\([\text{B2}]\)[/tex].
[tex]\[ \times \text{Equation} [\text{A2}] \rightarrow \text{Equation} [\text{B2}] \][/tex]

Therefore, the transformations are:
[tex]\[ \times \text{Equation} [\text{A1}] \rightarrow \text{Equation} [\text{B1}] \quad \text{(multiply by 1)} \][/tex]
[tex]\[ \times \text{Equation} [\text{A2}] \rightarrow \text{Equation} [\text{B2}] \quad \text{(multiply by 1)} \][/tex]

So the detailed transformation from System A to System B involves multiplying both equations by 1:
(a) 1, 1

### Transformation from System B to System C:

System B:
[tex]\[ \begin{cases} -3x + 3y = 6 \quad [\text{B1}] \\ x + 2y = 10 \quad [\text{B2}] \end{cases} \][/tex]

System C:
[tex]\[ \begin{cases} 9y = 36 \quad [\text{C1}] \\ x + 2y = 10 \quad [\text{C2}] \end{cases} \][/tex]

Given this, we now need to see how we transform each equation from System B to System C:

- From [tex]\([\text{B1}]\)[/tex] to [tex]\([\text{C1}]\)[/tex]:
- Notice that [tex]\([\text{C1}]\)[/tex] is obtained by multiplying [tex]\([\text{B1}]\)[/tex] by [tex]\(3\)[/tex] to get [tex]\(9y = 36\)[/tex].
[tex]\[ \times \text{Equation} [\text{B1}] \rightarrow \text{Equation} [\text{C1}] \][/tex]

- From [tex]\([\text{B2}]\)[/tex] to [tex]\([\text{C2}]\)[/tex]:
- We can see that [tex]\([\text{C2}]\)[/tex] is exactly the same as [tex]\([\text{B2}]\)[/tex]. Therefore, we multiply [tex]\([\text{B2}]\)[/tex] by 1.
[tex]\[ \times \text{Equation} [\text{B2}] \rightarrow \text{Equation} [\text{C2}] \][/tex]

So the detailed transformation from System B to System C involves:
(b) 3, 1

In summary:
[tex]\[ (a) \times \text{Equation} [\text{A1}] \rightarrow \text{Equation} [\text{B1}], \quad \times \text{Equation} [\text{A2}] \rightarrow \text{Equation} [\text{B2}] \quad = 1, 1 \][/tex]
[tex]\[ (b) \times \text{Equation} [\text{B1}] \rightarrow \text{Equation} [\text{C1}], \quad \times \text{Equation} [\text{B2}] \rightarrow \text{Equation} [\text{C2}] \quad = 3, 1 \][/tex]
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