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Ch. 1 - Equations and Inequalities
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Non è quello che usi tu?Cambia libro di testo
Capitolo 2, Problema 29

Exercises 27–40 contain linear equations with constants in denominators. Solve each equation. 20 - x/3 = x/2

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Identify the equation: \(20 - \frac{x}{3} = \frac{x}{2}\).
To eliminate the denominators, find the least common denominator (LCD) of 3 and 2, which is 6.
Multiply every term in the equation by 6 to clear the fractions: \(6 \times 20 - 6 \times \frac{x}{3} = 6 \times \frac{x}{2}\).
Simplify each term after multiplication: \(120 - 2x = 3x\).
Collect like terms by adding \$2x$ to both sides: \(120 = 3x + 2x\), then combine to get \(120 = 5x\). Finally, solve for $x$ by dividing both sides by 5.

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Solving Linear Equations

A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and a single variable. Solving such equations involves isolating the variable on one side to find its value. This often requires performing inverse operations like addition, subtraction, multiplication, or division.
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Solving Linear Equations with Fractions

Clearing Fractions by Multiplying Both Sides

When an equation contains fractions, multiplying both sides by the least common denominator (LCD) eliminates the denominators. This simplifies the equation into a linear form without fractions, making it easier to solve. For example, multiplying both sides by 6 clears denominators 2 and 3.
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Linear Inequalities with Fractions & Variables on Both Sides

Combining Like Terms and Simplifying

After clearing fractions, combine terms with the variable and constants on each side to simplify the equation. This step helps isolate the variable and reduces the equation to a simpler form, facilitating the final solution. Careful arithmetic ensures accuracy in solving.
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