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    Paul's Online Notes
    Home / Algebra Trig Review / Algebra / Solving Equations, Part II
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    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.

    August 27, 2020

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    Solving Equations, Part II

    Solve each of the following equations for \(y\).Show All Solutions Hide All Solutions

    1. \(\displaystyle x = \frac{{2y - 5}}{{6 - 7y}}\)
      Show Solution

      Here 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.

    2. \(3{x^2}\left( {3 - 5y} \right) + \sin x = 3xy + 8\)
      Show Solution

      This one solves the same way as the previous problem.

      \[\begin{align*}3{x^2}\left( {3 - 5y} \right) + \sin x & = 3xy + 8\\ 9{x^2} - 15{x^2}y + \sin x & = 3xy + 8\\ 9{x^2} + \sin x - 8 & = \left( {3x + 15{x^2}} \right)y\\ y & = \frac{{9{x^2} + \sin x - 8}}{{3x + 15{x^2}}}\end{align*}\]
    3. \(2{x^2} + 2{y^2} = 5\)
      Show Solution

      Same 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!

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