I apologize for the outage on the site yesterday and today. Lamar University is in Beaumont Texas and Hurricane Laura came through here and caused a brief power outage at Lamar. Things should be up and running at this point and (hopefully) will stay that way, at least until the next hurricane comes through here which seems to happen about once every 10-15 years. Note that I wouldn't be too suprised if there are brief outages over the next couple of days as they work to get everything back up and running properly. I apologize for the inconvienence.
Paul
August 27, 2020
Solving Equations, Part II
Solve each of the following equations for \(y\).Show All Solutions Hide All Solutions
- \(\displaystyle x = \frac{{2y - 5}}{{6 - 7y}}\)
Show SolutionHere all we need to do is get all the \(y\)’s on one side, factor a \(y\) out and then divide by the coefficient of the \(y\)
\[\begin{align*}x & = \frac{{2y - 5}}{{6 - 7y}}\\ x\left( {6 - 7y} \right) & = 2y - 5\\ 6x - 7xy & = 2y - 5\\ 6x + 5 & = \left( {7x + 2} \right)y\\ y & = \frac{{6x + 5}}{{7x + 2}}\end{align*}\]Solving equations for one of the variables in it is something that you’ll be doing on occasion in a Calculus class so make sure that you can do it.
- \(3{x^2}\left( {3 - 5y} \right) + \sin x = 3xy + 8\)
Show SolutionThis one solves the same way as the previous problem.
\[\begin{align*}3{x^2}\left( {3 - 5y} \right) + \sin x & = 3xy + 8\\ 9{x^2} - 15{x^2}y + \sin x & = 3xy + 8\\ 9{x^2} + \sin x - 8 & = \left( {3x + 15{x^2}} \right)y\\ y & = \frac{{9{x^2} + \sin x - 8}}{{3x + 15{x^2}}}\end{align*}\] - \(2{x^2} + 2{y^2} = 5\)
Show SolutionSame thing, just be careful with the last step.
\[\begin{align*}2{x^2} + 2{y^2} & = 5\\ 2{y^2} & = 5 - 2{x^2}\\ {y^2} & = \frac{1}{2}\left( {5 - 2{x^2}} \right)\\ y & = \pm \sqrt {\frac{5}{2} - {x^2}} \end{align*}\]Don’t forget the “\( \pm \)” in the solution!