How To Solve Equations For The Indicated Variable

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.

1. Isolate the Variable

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

\(4x^2 = 7\)

2. Divide by the Coefficient (If Present)

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

\(x^2 = 1.75\)

3. Take the Root of the Equation

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

\(x = 1.32\)

1. Isolate the Variable Expression

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:

\(\sqrt{x + 27} + 11 – 11 = 15 – 11 = 4\)

2. Apply an Exponent to Both Sides of the Equation

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

\(x + 27 = 16\)

3. Cancel the Constant

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:

\(x = 16 – 27 = -11\)

1. Set the Quadratic Equation Equal to Zero

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

\(2x^2 – x – 1 = 0\)

2. Factor or Complete the Square

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:

\(2x^2 – x – 1 = 0 \text{ becomes } (2x + 1)(x – 1) = 0\)

3. Solve for the Variable

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.

1. Factor the Denominators

Factor each denominator. For example,

\(\frac{1}{x – 3} + \frac{1}{x + 3} = \frac{10}{x^2 – 9}\)

can be factored to become:

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

2. Multiply by Least Common Multiple of Denominators

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

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

3. Cancel and Solve for the Variable

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

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

1. Isolate the Exponential Expression

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

\(100×(14^x) + 6 = 10\)

becomes

\(\begin{aligned}\)
\(100×(14^x) + 6 – 6 &= 10 – 6\)
\(&= 4\)
\(\end{aligned}\)

2. Cancel the Coefficient

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

\(100×(14^x) = 4\)

becomes

\(\frac{100×(14^x)}{100} = \frac{4}{100} \
\,\)
\(14^x = 0.04\)

3. Use the Natural Logarithm

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):

\(\ln(14^x)= \ln(0.04)\)
\(x × \ln(14) = \ln\bigg(\frac{1}{25}\bigg)\)
\(x × \ln(14) = \ln(1) – \ln(25)\)
\(x × \ln(14) = 0 – \ln(25)\)

4. Solve for the Variable

Solve the equation for the variable. For example,

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

1. Isolate the Logarithmic Expression

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

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

2. Apply an Exponent

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:

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

3. Solve for the Variable

Solve the equation for the variable. For example,

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

becomes

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