To create a true equivalent system for the equations provided for System A and System B, we first need to analyze what we have.
For System A, we have the equation:
- \(-x + y = 1\)
For System B, we have the equations:
- \(x + y = 11\)
- \(x + y = 11\) (which is the same as the first equation)
To find an equivalent equation to \( (-x + y = 1\), we can manipulate it to explore its relationships.
We can start by rearranging \( -x + y = 1\) to express \(y\) in terms of \(x\): \[ y = x + 1 \]
Next, we can manipulate System B's equations. Both equations in System B state that: \[ x + y = 11 \]
We can solve for \(y\) in terms of \(x\) from System B as well: \[ y = 11 - x \]
To create a true equivalent system, we can set the two derived equations for \(y\) equal to each other: \[ x + 1 = 11 - x \]
Now, we can solve for \(x\): \[ x + x = 11 - 1 \] \[ 2x = 10 \] \[ x = 5 \]
Now, substituting \(x = 5\) back into one of the equations (e.g., \(y = x + 1\)): \[ y = 5 + 1 = 6 \]
Thus, a valid system of equations that is equivalent might be:
- \(x = 5\)
- \(y = 6\)
Alternatively, if we want to create another equivalent equation, we can express the first equation in terms of \(y\): Multiplying the original equation of System A by -1: \[ x - y = -1 \]
This can be one of the equations in the equivalent system.
So, one example of an equivalent system for System A could be:
- \(x - y = -1\) (equivalent of \(-x + y = 1\))
- The second equation still can remain the same \(x + y = 11\).
In conclusion, an equivalent system could be:
- \(x - y = -1\)
- \(x + y = 11\)
Any modification or rearrangement of these equations maintaining the equivalency would also be valid.
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