Question
Answer two questions about Systems \( A \) and 15 : \[ \begin{array}{cc} \text { System } A & \text { System } B \\ \left\{\begin{array}{l} 6 x-5 y-1 \\ -2 x+2 y-1 \end{array}\right. & \left\{\begin{array}{c} 4 x-3 y=0 \\ -2 x+2 y=-1 \end{array}\right. \end{array} \] 1) How can we get System \( B \) from System \( A \) ? Choose 1 answer: (A) Replace one equation with the sum/difference of both equations (B) Replace only the left-hand side of one equation with the sum/difference of the left-hand sides of both equations (c) Replace onc equation with a multiple of itself (D) Replace one equation with a multiple of the other equation
Ask by Salazar Simmons. in the United States
Feb 26,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
To get System B from System A, we add both equations of System A together. This results in the first equation of System B. Therefore, the correct choice is:
(A) Replace one equation with the sum/difference of both equations.
Solution
We first note that if we write System A in standard form the equations become
(1) 6x – 5y = 1
(2) –2x + 2y = –1
Now, if we add equations (1) and (2) of System A, we have
(6x – 5y) + (–2x + 2y) = 1 + (–1)
=> 4x – 3y = 0
This is exactly the first equation of System B. The second equation in System B is identical to equation (2) of System A, i.e., –2x + 2y = –1.
Thus System B was obtained by replacing the first equation of System A with the sum of both equations.
Therefore, the correct answer is
(A) Replace one equation with the sum/difference of both equations.
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Simplify this solution The Deep Dive
To get System \( B \) from System \( A \), you can replace the first equation of System \( A \) with the sum of both equations. By doing this, the first equation \( 6x - 5y - 1 \) becomes \( 4x - 3y = 0 \) when you added \( -2x + 2y - 1 \) to it, which matches the first equation in System \( B \). So the answer is (A). Furthermore, in System \( B \), the second equation is just the second equation of System \( A \) multiplied by \(-1\), which is a common technique to eliminate variables during calculations. This keeps consistency in the solution set while simplifying the equations, highlighting that both systems represent the same lines on a graph.
