At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
\[y = (x-1)/(2x+3)\]
I am attempting to solve for x.
But I find myself going in a circle instead of isolating x. Can you give me a hint?
Multiply both sides by 2x+3, you should get (2x+3)y=x-1 2xy+3y=x-1 2xy+3y+1=x 2xy-x=-3y-1 x(2y-1)=-3y-1 x=(-3y-1)/(2y-1)
Try factoring out - x instead, which will make for a positive numerator, and I like that better. Thanks for helping me get unstuck.
Your welcome, how is that calc group doing?
I never connected with anybody at my ideal time. I did finish my Calc 1 class locally, and just found out I got an A on the final.
Thank you for remembering.
I don't think I could have gotten an A at MIT, though. I am working through the OCW class on my own in prep for Calc 2 next semester. The teaching is better, the recitations are more logically connected to the future of calculus work coming in the class, and the problems are clearly better structured to teach something specific while doing the homework as a side benefit to practicing a particular kind of problem. That makes the OCW homework feel much much more rewarding.