How to Solve Equations for the Indicated Variable

Solving variable equations is the essence of algebra.
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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

    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

    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

    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

    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

    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

    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

    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

    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

    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

    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)}

    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)

    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

    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}

    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

    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)

    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

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

    2\ln(3x) = 4 \text{ becomes } \ln(3x) = \frac{4}{2} = 2

    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

    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

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