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Section 1.11 : Solving Equations, Part II

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

1. \(\displaystyle x = \frac{{2y - 5}}{{6 - 7y}}\)
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   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. \(3{x^2}\left( {3 - 5y} \right) + \sin x = 3xy + 8\)
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   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. \(2{x^2} + 2{y^2} = 5\)
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   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!