Question

Which of the following shows the equation that creates a true equivalent system?
System A | System B
-x + y = 1 | x + y = 11
x + y = 11 | BLANK

Answer 1

dude just give an equation that fills in the BLANK

Answer 2

To determine a true equivalent system to System A (consisting of the equations \(-x + y = 1\) and \(x + y = 11\)), we can manipulate the equations to find an equivalent form for System B.

From System A, we have the first equation:

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

To find an equivalent equation, we can solve this equation for \(y\):

\[
y = x + 1
\]

Next, we can substitute \(y\) in the second equation of System A:

2) \(x + (x + 1) = 11\)

This simplifies to:

\[
2x + 1 = 11
\]

Now, subtract 1 from both sides:

\[
2x = 10
\]

And then divide by 2:

\[
x = 5
\]

Now substitute \(x = 5\) back into the equation for \(y\):

\[
y = 5 + 1 = 6
\]

So the point \((5, 6)\) will satisfy both of the original equations in System A.

Now, we can form a new equivalent equation for System B that goes directly from these results. Since we obtained \(y = x + 1\) from manipulating System A, we can express our System B such that both parts will be satisfied by the variable values derived.

Thus, the second equation for System B could also be a rearrangement, or manipulation of existing equations. An option for a compatible equation can include substituting variables:

The equivalent equation for System B can be:
\[
y - 6 = 0
\]

This indicates any point with \(y\) equal to 6 will satisfy both equations.

So a valid equivalent system could be expressed as the following:

**System A**:
1) \(-x + y = 1\)
2) \(x + y = 11\)

**System B**:
1) \(x + y = 11\)
2) \(y = 6\) (or \(y - 6 = 0\))

Thus, both systems will represent the same relationship between \(x\) and \(y\).

Answer 3

I apologize for the confusion! To fill in the blank for System B, we can use the first equation from System A and rewrite it in terms of \(y\).

The equation from System A is:
\(-x + y = 1\)

We can rearrange it as follows:
\[ y = x + 1 \]

So, you can fill in the blank in System B with:
**\(y = x + 1\)**

Now, System B will be:

System B:
- \(x + y = 11\)
- \(y = x + 1\)

This provides an equivalent system that matches System A.