To determine how we can transform System A into System B, let's carefully analyze the given systems:
System A:
[tex]\[
\begin{cases}
x + 3y = -9 \\
2x + y = 4
\end{cases}
\][/tex]
System B:
[tex]\[
\begin{cases}
3x + 4y = -9 \\
2x + y = 4
\end{cases}
\][/tex]
Notice that both systems share the second equation, [tex]\(2x + y = 4\)[/tex]. So, the transformation must involve changing the first equation of System A, [tex]\(x + 3y = -9\)[/tex], into the first equation of System B, [tex]\(3x + 4y = -9\)[/tex].
Let's explore each option to find which one is correct:
Option (A): Replace one equation with the sum/difference of both equations
Let's consider this option: we'll try to combine the two equations in System A to form a new equation.
1. Multiply the second equation:
[tex]\[
2x + y = 4
\][/tex]
by some factor, e.g., 2:
[tex]\[
2(2x + y) = 2 \cdot 4 \\
4x + 2y = 8
\][/tex]
2. Add this to the first equation:
[tex]\[
x + 3y = -9 \\
\][/tex]
[tex]\[
(x + 3y) + (4x + 2y) = -9 + 8 \\
x + 3y + 4x + 2y = -9 + 8 \\
5x + 5y = -1
\][/tex]
This approach doesn’t lead us directly to [tex]\(3x + 4y = -9\)[/tex]. Let's try another.
Option (B): Replace only the left-hand side of one equation with the sum/difference of the left-hand sides of both equations
Let’s consider combining just the left-hand sides of the two equations in System A:
1. Adding the left-hand sides:
[tex]\[
(x + 3y) + (2x + y) = x + 3y + 2x + y = 3x + 4y \\
\][/tex]
Now, we place [tex]\(3x + 4y\)[/tex] as the new left-hand side without changing the right-hand side to:
1. The right-hand sides combine:
[tex]\[
-9 + 4 \neq -9\\
\][/tex]
So this approach alters the right-hand side combined, which doesn't match the first equation in System B.
Option (C): Replace one equation with a multiple of itself
Let's consider multiplying the first equation by a constant factor:
[tex]\[
n(x + 3y) = n(-9)
\][/tex]
Choose [tex]\(n = 3:\)[/tex]
[tex]\[
3(x + 3y) = 3(-9)\\
=3x+9y=-27
\][/tex]
This does not match the first equation in System B [tex]\( (3x + 4y = -9)\)[/tex].
Option (D): Replace one equation with a multiple of the other equation
Let's determine if a multiple of the second equation equals the first equation in System B:
\[
n(2x + y) = n(4)
3x + 6y \quad n = 0.5 \text{ not 4}
=3x + 4y \neq -9
]
None multiplied.
From exploring all options, Option (A): Replace one equation with the sum/difference of both equations, allows suitable transformations of System A to System B:
\textbf{Answer}:
(A) Replace one equation with the sum/difference of both equations
