Question about an AoPS problem

[swimmermom3]

I have finally had a chance to sit down and start doing my homework in the AoPS Intemdiate Algebra book. I am only on problem 1.5 on p. 2 and I have a question - or three. :tongue_smilie:

 

I had no problem solving part a, but the explanations for parts b-d are escaping me. On p. 5 with the explanation, I am fine up to the point of "We cannot have c = 3/2." I understand why one can't divide by 0, but where does one come up with c=3/2 or x = 1/2 as possible options.

 

I sincerely hope that I am forgetful (did this 30 years ago?) and not dense. Sigh.

 

I always want to do everything my student is doing, especially if I am responsible for teaching and the grade. I am just not sure what adding AoPS to the line-up with chemistry and AP Euro is going to do except bend my mind slowly - in a good way of course. :D

[regentrude]

I had no problem solving part a, but the explanations for parts b-d are escaping me. On p. 5 with the explanation, I am fine up to the point of "We cannot have c = 3/2." I understand why one can't divide by 0, but where does one come up with c=3/2 or x = 1/2 as possible options.

 

If I take the original equation, it contains a term 1/x, which is not defined for x=0. Anything I do when I rearrange the equation will only be valid if x is not zero.

The original equation is also undefined if the entire denominator is zero: 2 - 1/x =0. Solving this equation gives x= 1/2. So, when I rearrange the equation for x, it is valid only if x is not 1/2 either.

Now, the rearranged equation for x also has a denominator that must not be zero either. Solving 2c-3 =0 gives the c=3/2. For c=3/2 the rearranged equation is not defined.

 

So, the values of c for which there are no solutions x must be c=3/2, and whatever values of c lead to x=0 and x=1/2, respectively.

Setting x=c/(2c-3) equal to zero leads to c=0.

Setting x=c/(2c-3) equal to 1/2 leads to 2c=2c-3 which is impossible; no value of c will lead to x=1/2.

 

Thus, c=0 and c=3/2 do not have a solution for x. Let's check this by putting these values into the original equation:

we see that obviously c can never be zero since the left side is not zero for any value of x, and that c=3/2 would require 1/x to be equal to zero which can not be possible either.

[swimmermom3]

If I take the original equation, it contains a term 1/x, which is not defined for x=0. Anything I do when I rearrange the equation will only be valid if x is not zero.

The original equation is also undefined if the entire denominator is zero: 2 - 1/x =0. Solving this equation gives x= 1/2. So, when I rearrange the equation for x, it is valid only if x is not 1/2 either.

Now, the rearranged equation for x also has a denominator that must not be zero either. Solving 2c-3 =0 gives the c=3/2. For c=3/2 the rearranged equation is not defined.

 

So, the values of c for which there are no solutions x must be c=3/2, and whatever values of c lead to x=0 and x=1/2, respectively.

Setting x=c/(2c-3) equal to zero leads to c=0.

Setting x=c/(2c-3) equal to 1/2 leads to 2c=2c-3 which is impossible; no value of c will lead to x=1/2.

 

Thus, c=0 and c=3/2 do not have a solution for x. Let's check this by putting these values into the original equation:

we see that obviously c can never be zero since the left side is not zero for any value of x, and that c=3/2 would require 1/x to be equal to zero which can not be possible either.

 

Thank you so much for the explanation.

 

AoPS seems like it goes deeper than anything we've done here. I wish we had started  a bit earlier.