Question

Use the table to answer the question.

System A System B
−x+y=1

x+y=11

x+y=11

?

Which of the following shows the equation that creates a true equivalent system?

(1 point)

Answer 1

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:
1. \(-x + y = 1\)

For System B, we have the equations:
1. \(x + y = 11\)
2. \(x + y = 11\) (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:
1. \(x = 5\)
2. \(y = 6\)

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:

1. \(x - y = -1\) (equivalent of \(-x + y = 1\))
2. The second equation still can remain the same \(x + y = 11\).

In conclusion, an equivalent system could be:
1. \(x - y = -1\)
2. \(x + y = 11\)

Any modification or rearrangement of these equations maintaining the equivalency would also be valid.