Skip to main content
Workforce LibreTexts

3.2: Solving a Formula for a Variable (Rewriting a Formula)

  • Page ID
    20036
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    In the Water Industry, we use lots of formulas for computing area, volume, pressure, cost, and so much more. Sometimes, we need to rewrite a formula in terms of a different quantity.

    When solving a formula for a particular variable, it is sometimes helpful to place a rectangle or a circle around the variable you are isolating since formulas contain many variables and we may lose our focus as to which variable we are isolating. We will proceed as we did for solving linear equations by “undoing” operations to isolate the variable.

    Example \(\PageIndex{1}\)

    The volume of a rectangular tank is given by \(V=LWH\), rewrite the formula by solving for \(H\).

    Solution

    To “undo” the multiplication of H by LW, we divide both sides of the equation by LW to obtain:

    \[\frac{V}{ \boldsymbol{LW}}=\frac{LWH}{ \boldsymbol{ LW}}\]

    \[{V}\over {LW}=H\]

    Example \(\PageIndex{2}\)

    The filtration rate is defined as Filtration Rate \(=\frac{Flow}{Surface \, \, Area}\). Solve this equation for the Surface Area.

    Solution

    Step 1: Multiply both sides of the equation by Surface Area (to remove the Surface Area from the denominator. In other words, we cannot isolate Surface Area until it is no longer in the denominator)

    \[\left(Filtration\,\, Rate\right)\cdot \left(\boldsymbol{Surface\,\, Area}\right) =\frac{Flow}{Surface\,\, Area}\cdot \left(\boldsymbol{Surface\,\, Area}\right)\]

    \[\left(Filtration\,\, Rate\right) \cdot \left(Surface\,\, Area\right) =Flow\]

    Step 2: Isolate the Surface Area by dividing both sides by Filtration Rate

    \[\frac{\left( Filtration\,\, Rate\right)\cdot \left(Surface\,\, Area\right) }{(\boldsymbol{Filtration\,\, Rate})}=\frac{Flow}{(\boldsymbol{Filtration\,\, Rate})} \]

    \[Surface\,\, Area=\frac{Flow}{Filtration\,\, Rate}\]

    This is helpful if we need to calculate the Surface Area and we are given the flow and the filtration rate.


    This page titled 3.2: Solving a Formula for a Variable (Rewriting a Formula) is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Kelly Brooks.