Skip to main content
Workforce LibreTexts

4.1: Solving Equations

  • Page ID
    7091
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    Equations are a vital component to solve many mathematical questions. In fact most of all water related mathematical questions would require the use of one if not several equations. An equation is a mathematical statement that both sides of an “equal sign” are in fact equal.

    Example \(\PageIndex{1}\)

    \(3 + 1 = 4\) or \(5 - 3 = 4 - 2\)

    Mathematical equations involve solving for an unknown or a variable. A variable is typically represented with a letter. That “letter” is what you are trying to solve.

    Example \(\PageIndex{2}\)

    \(3+X=4\)

    What number must X be in order to make this equation correct?

    Obviously, the answer is 1.

    Now, this is a very simple example, but it will allow you to begin the process for solving more complex equations.

    How did you solve the example above? Did you simply look at the equation and intuitively know that the answer was 1? Or did you use some mathematical process to solve for the variable? Let’s break it down.

    Steps to solving equations;

    1. Get the unknown or unknowns on one side of the equation.
    2. Get the numbers on the other side of the equation.
    3. Try and follow these helpful tips.
      1. If the problem is an addition problem, you will need to subtract
      2. If the problem is a subtraction problem, you will need to add
      3. If the problem is a multiplication problem, you will need to divide
      4. If the problem is a division problem, you will need to multiply

    Depending on the type of problem, this will involve adding, subtracting, multiplying, or dividing. Perhaps several steps will need to be done to solve for the variable.

    Since we will not be adding or subtracting in this course, we will focus on multiplication and division problems.

    If the variable is next to a fraction or directly next to a number, solving for the variable will require either multiplying or dividing.

    These next two examples are using multiplication and division to get the variable by itself. In order to prevent confusion, the variable of choice will be something other than an “X.”

    Example \(\PageIndex{3}\)

    \(2N=10\) What multiplied by 2 equals 10? In order to solve for “N” you will need to divide both sides of the equation by 2.
    \(\dfrac{2N}{2}=10\) By dividing the left side of the equation by 2, you isolate the variable. However, since you divided one side of the equation by a number you must do the same to the other side of the equation.
    \(N=\dfrac{10}{2}\)
    \(N=5\)

    In this next example, a fraction of a variable will equal a number.

    Example \(\PageIndex{4}\)

    \(\dfrac{2}{3} N=14\) In order to isolate the variable in this type of problem, you will need to multiply and divided. First, divide both sides of the equation by 2 and then multiply both sides of the equation by 3.
    \(\dfrac{2}{(2) 3} N=\dfrac{14}{2}\) The two cancels out on the left side of the equation leaving N over 3 and 14 divided by 2 becomes 7.
    \(\dfrac{N}{3}=7\) Now, multiply both sides of the equation by 3.
    \(\dfrac{(3) N}{3}=7 \times 3\) Here, the 3 cancels out on the left and the right side of the equation becomes 21, which is the answer.
    \(N=21\)

    How can you prove (or show) that 21 is the correct answer in the above example.

    Exercise 4.1

    1. \(2 X=10\)
    2. \(3 N=15\)
    3. \(\dfrac{3}{4} G=9\)
    4. \(4 H+2=10\)
    5. \(\dfrac{1}{5} J+4=5\)
    6. \(13=\dfrac{3}{6} K-1\)
    7. \(3 B-5=2 B+8\)
    8. \(10 C-7=3 C+14\)
    9. \(\dfrac{4}{3} S-3=5\)
    10. \(14=2F+2\)
    11. \(7 X+2=9 X-10\)
    12. \(15=5+\dfrac{2}{5} P\)

    4.1: Solving Equations is shared under a CC BY license and was authored, remixed, and/or curated by LibreTexts.

    • Was this article helpful?