How to Solve Equations for the Indicated Variable

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Elementary algebra is one of the main branches of mathematics. Algebra introduces the concept of using variables to represent numbers and defines the rules on how to manipulate equations containing these variables. Variables are important because they allow for the formulation of generalized mathematical laws and allow the introduction of unknown numbers into equations. It is these unknown numbers that are the focus of algebra problems, which usually prompt you to solve for the indicated variable. The "standard" variables in algebra are frequently represented as x and y.

Solving Linear and Parabolic Equations

  1. Isolate the Variable

  2. Move any constant values from the side of the equation with the variable to the other side of the equals sign. For example, for the equation

    4x^2 + 9 = 16

    subtract 9 from both sides of the equation to remove the 9 from the variable side:

    4x^2 + 9 - 9 = 16 - 9

    which simplifies to

    4x^2 = 7
  3. Divide by the Coefficient (If Present)

  4. Divide the equation by the coefficient of the variable term. For example,

    \text{if } 4x^2 = 7 \text{ then } \frac{4x^2}{4} = \frac{7}{4}

    which results in

    x^2 = 1.75
  5. Take the Root of the Equation

  6. Take the proper root of the equation to remove the exponent of the variable. For example,

    \text{if } x^2 = 1.75 \text{ then } \sqrt{x^2} = \sqrt{1.75}

    which results in

    x = 1.32

Solve for the Indicated Variable With Radicals

  1. Isolate the Variable Expression

  2. Isolate the expression containing the variable by using the appropriate arithmetic method to cancel out the constant on the side of the variable. For example, if

    \sqrt{x + 27} + 11 = 15

    you would isolate the variable using subtraction:

    \sqrt{x + 27} + 11 - 11 = 15 - 11 = 4
  3. Apply an Exponent to Both Sides of the Equation

  4. Raise both sides of the equation to the power of the root of the variable to rid the variable of the root. For example,

    \sqrt{x + 27} = 4 \text{ then } (\sqrt{x + 27})^2 = 4^2

    which gives you

    x + 27 = 16
  5. Cancel the Constant

  6. Isolate the variable by using the appropriate arithmetic method to cancel out the constant on the side of the variable. For example, if

    x + 27 = 16

    by using subtraction:

    x = 16 - 27 = -11

Solving Quadratic Equations

  1. Set the Quadratic Equation Equal to Zero

  2. Set the equation equal to zero. For example, for the equation

    2x^2 - x = 1

    subtract 1 from both sides to set the equation to zero

    2x^2 - x - 1 = 0
  3. Factor or Complete the Square

  4. Factor or complete the square of the quadratic, whichever is easier. For example, for the equation

    2x^2 - x - 1 = 0

    it is easiest to factor so:

    2x^2 - x - 1 = 0 \text{ becomes } (2x + 1)(x - 1) = 0
  5. Solve for the Variable

  6. Solve the equation for the variable. For example, if

    (2x + 1)(x - 1) = 0

    then the equation equals zero when:

    2x + 1 = 0

    Implies that

    2x = -1 \text{, so } x = -\frac{1}{2}

    or when

    \text{when } x - 1 = 0\text{, you get } x = 1

    These are the solutions to the quadratic equation.

An Equation Solver for Fractions

  1. Factor the Denominators

  2. Factor each denominator. For example,

    \frac{1}{x - 3} + \frac{1}{x + 3} = \frac{10}{x^2 - 9}

    can be factored to become:

    \frac{1}{x - 3} + \frac{1}{x + 3} = \frac{10}{(x - 3)(x + 3)}
  3. Multiply by Least Common Multiple of Denominators

  4. Multiply each side of the equation by the least common multiple of the denominators. The least common multiple is the expression that each denominator can divide evenly into. For the equation

    \frac{1}{x - 3} + \frac{1}{x + 3} = \frac{10}{(x - 3)(x + 3)}

    the least common multiple is (​x​ − 3)(​x​ + 3). So,

    (x - 3)(x + 3) \bigg(\frac{1}{x - 3} + \frac{1}{x + 3}\bigg) = (x - 3)(x + 3)\bigg(\frac{10}{(x - 3)(x + 3)}\bigg)

    becomes

    \frac{(x - 3)(x + 3)}{x - 3} + \frac{(x - 3)(x + 3)}{x + 3} = (x - 3)(x + 3)\bigg(\frac{10}{(x - 3)(x + 3)}\bigg)
  5. Cancel and Solve for the Variable

  6. Cancel terms and solve for ​x​. For example, cancelling terms for the equation

    \frac{(x - 3)(x + 3)}{x - 3} + \frac{(x - 3)(x + 3)}{x + 3} = (x - 3)(x + 3)\bigg(\frac{10}{(x - 3)(x + 3)}\bigg)

    gives:

    (x + 3) + (x - 3) = 10

    Leads to

    2x = 10 \text{, and } x = 5

Dealing With Exponential Equations

  1. Isolate the Exponential Expression

  2. Isolate the exponential expression by cancelling any constant terms. For example,

    100×(14^x) + 6 = 10

    becomes

    \begin{aligned} 100×(14^x) + 6 - 6 &= 10 - 6 \\ &= 4 \end{aligned}
  3. Cancel the Coefficient

  4. Cancel out the coefficient of the variable by dividing both sides by the coefficient. For example,

    100×(14^x) = 4

    becomes

    \frac{100×(14^x)}{100} = \frac{4}{100} \\ \,\\ 14^x = 0.04
  5. Use the Natural Logarithm

  6. Take the natural log of the equation to bring down the exponent containing the variable. For example,

    14^x = 0.04

    can be written as (using some properties of logarithms):

    \ln(14^x)= \ln(0.04) \\ x × \ln(14) = \ln\bigg(\frac{1}{25}\bigg) \\ x × \ln(14) = \ln(1) - \ln(25) \\ x × \ln(14) = 0 - \ln(25)
  7. Solve for the Variable

  8. Solve the equation for the variable. For example,

    x × \ln(14) = 0 - \ln(25) \text{ becomes } x = \frac{-\ln(25)}{\ln(14)} = -1.22

A Solution for Logarithmic Equations

  1. Isolate the Logarithmic Expression

  2. Isolate the natural log of the variable. For example, the equation

    2\ln(3x) = 4 \text{ becomes } \ln(3x) = \frac{4}{2} = 2
  3. Apply an Exponent

  4. Convert the log equation to an exponential equation by raising the log to an exponent of the appropriate base. For example,

    \ln(3x) = 2

    becomes:

    e^{\ln(3x)}= e^2
  5. Solve for the Variable

  6. Solve the equation for the variable. For example,

    e^{\ln(3x)}= e^2

    becomes

    \frac{3x}{3} = \frac{e^2}{3} \text{ so } x = 2.46

References

About the Author

Luc Braybury began writing professionally in 2010. He specializes in science and technology writing and has published on various websites. He received his Bachelor of Science in applied physics from Armstrong Atlantic State University in Savannah, Ga.

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