Problem:
Solution:
For part (a), the second equation seems to be simpler and therefore we will start there. Assuming x≠0, we can write y=1x. Putting this back to the first equation gives:
x2+1x2−1=0.
This works if x≠0, but what if x=0? Turn out we do not need to worry about it because if x=0, then the second equation cannot possibly be satisfied. Envisioning part (b), let's convert it into a polynomial by multiplying through x2 and get:
x4−x2+1=0
For part (b), we use the hint and get (xy−1)(xy+1)=x2y2−1. To eliminate y2, we multiply the first equation by x2 so we get x2(x2+y2−1)=x4+x2y2−x2.
Finally we get x4−x2+1=x2(x2+y2−1)−(xy−1)(xy+1), therefore x4−x2+1∈⟨x2+y2−1,xy−1⟩
Solution:
For part (a), the second equation seems to be simpler and therefore we will start there. Assuming x≠0, we can write y=1x. Putting this back to the first equation gives:
x2+1x2−1=0.
This works if x≠0, but what if x=0? Turn out we do not need to worry about it because if x=0, then the second equation cannot possibly be satisfied. Envisioning part (b), let's convert it into a polynomial by multiplying through x2 and get:
x4−x2+1=0
For part (b), we use the hint and get (xy−1)(xy+1)=x2y2−1. To eliminate y2, we multiply the first equation by x2 so we get x2(x2+y2−1)=x4+x2y2−x2.
Finally we get x4−x2+1=x2(x2+y2−1)−(xy−1)(xy+1), therefore x4−x2+1∈⟨x2+y2−1,xy−1⟩
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