Struggling with making a the subject of abd = 1/c^2 + ac; not sure if I am on the right track?

[HATLEY1997]

 

[stapel]

How did you get from:

[imath]\qquad abc^2 - ac^3 = 1[/imath]

...which is correct, to:

[imath]\qquad a - ac^3 = \dfrac{1}{bc^2}[/imath]

...? How did you factor the [imath]bc^2[/imath] out of the [imath]ac^3[/imath], in order to divide through by it?

 

[Steven G]

10 - 4= 6

Is 10/2 - 4 = 6/2?? That is can you divide the whole right side by 2 but only part of the left hand side by 2. That is what you are doing in your work.

ab = 1/c² + ac
ab-ac = 1/c²
a(b-c) = 1/c²
Can you finish from here?

 

[HATLEY1997]

How did you get from:

[imath]\qquad abc^2 - ac^3 = 1[/imath]

...which is correct, to:

[imath]\qquad a - ac^3 = \dfrac{1}{bc^2}[/imath]

...? How did you factor the [imath]bc^2[/imath] out of the [imath]ac^3[/imath], in order to divide through by it?

Am I on the right track here? I have went back and had another look

 

[Steven G]

What you did was fine. I would have factored out the c²in the denominator.

I also feel that what you did was a bit more than necessary. Here is what I would have done. In the end you solved it and that is what really matters.

ab = 1/c² + ac

ab-ac = 1/c²

a(b-c) = 1/c²

a= 1/[c²(b-c)]

 

[HATLEY1997]

What you did was fine. I would have factored out the c²in the denominator.

I also feel that what you did was a bit more than necessary. Here is what I would have done. In the end you solved it and that is what really matters.

ab = 1/c² + ac

ab-ac = 1/c²

a(b-c) = 1/c²

a= 1/[c²(b-c)]

thanks for that really helpful