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:
- \(-x + y = 1\)
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:
- \(x + (x + 1) = 11\)
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:
- \(-x + y = 1\)
- \(x + y = 11\)
System B:
- \(x + y = 11\)
- \(y = 6\) (or \(y - 6 = 0\))
Thus, both systems will represent the same relationship between \(x\) and \(y\).
1 answer