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Solving an equation in one variable

Solve the following equation: 2 x + 7 = 19.

This equation can be written in the form a x + b = 0 by subtracting 19 from both sides. However, we may proceed to solve the equation in its original form by performing algebraic operations.

2 x + 7 = 19 2 x = 12 Subtract 7 from both sides . x = 6 Multiply both sides by  1 2  or divide by 2 .

The solution is x = 6.

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Solve the linear equation in one variable: 2 x + 1 = −9.

x = −5

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Solving an equation algebraically when the variable appears on both sides

Solve the following equation: 4 ( x −3 ) + 12 = 15 −5 ( x + 6 ) .

Apply standard algebraic properties.

4 ( x 3 ) + 12 = 15 5 ( x + 6 ) 4 x 12 + 12 = 15 5 x 30 Apply the distributive property . 4 x = −15 5 x Combine like terms . 9 x = −15 Place  x terms on one side and simplify . x = 15 9 Multiply both sides by  1 9 , the reciprocal of 9 . x = 5 3
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Solve the equation in one variable: −2 ( 3 x 1 ) + x = 14 x .

x = −3

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Solving a rational equation

In this section, we look at rational equations that, after some manipulation, result in a linear equation. If an equation contains at least one rational expression, it is a considered a rational equation .

Recall that a rational number is the ratio of two numbers, such as 2 3 or 7 2 . A rational expression    is the ratio, or quotient, of two polynomials. Here are three examples.

x + 1 x 2 4 , 1 x 3 , or 4 x 2 + x 2

Rational equations have a variable in the denominator in at least one of the terms. Our goal is to perform algebraic operations so that the variables appear in the numerator. In fact, we will eliminate all denominators by multiplying both sides of the equation by the least common denominator    (LCD).

Finding the LCD is identifying an expression that contains the highest power of all of the factors in all of the denominators. We do this because when the equation is multiplied by the LCD, the common factors in the LCD and in each denominator will equal one and will cancel out.

Solving a rational equation

Solve the rational equation: 7 2 x 5 3 x = 22 3 .

We have three denominators; 2 x , 3 x , and 3. The LCD must contain 2 x , 3 x , and 3. An LCD of 6 x contains all three denominators. In other words, each denominator can be divided evenly into the LCD. Next, multiply both sides of the equation by the LCD 6 x .

( 6 x ) [ 7 2 x 5 3 x ] = [ 22 3 ] ( 6 x ) ( 6 x ) ( 7 2 x ) ( 6 x ) ( 5 3 x ) = ( 22 3 ) ( 6 x ) Use the distributive property . ( 6 x ) ( 7 2 x ) ( 6 x ) ( 5 3 x ) = ( 22 3 ) ( 6 x ) Cancel out the common factors . 3 ( 7 ) 2 ( 5 ) = 22 ( 2 x ) Multiply remaining factors by each numerator . 21 10 = 44 x 11 = 44 x 11 44 = x 1 4 = x
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A common mistake made when solving rational equations involves finding the LCD when one of the denominators is a binomial—two terms added or subtracted—such as ( x + 1 ) . Always consider a binomial as an individual factor—the terms cannot be separated. For example, suppose a problem has three terms and the denominators are x , x 1 , and 3 x 3. First, factor all denominators. We then have x , ( x 1 ) , and 3 ( x 1 ) as the denominators. (Note the parentheses placed around the second denominator.) Only the last two denominators have a common factor of ( x 1 ) . The x in the first denominator is separate from the x in the ( x 1 ) denominators. An effective way to remember this is to write factored and binomial denominators in parentheses, and consider each parentheses as a separate unit or a separate factor. The LCD in this instance is found by multiplying together the x , one factor of ( x 1 ) , and the 3. Thus, the LCD is the following:

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Source:  OpenStax, College algebra. OpenStax CNX. Feb 06, 2015 Download for free at https://legacy.cnx.org/content/col11759/1.3
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