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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 50

Exercises 41–60 contain rational equations with variables in denominators. For each equation, a. write the value or values of the variable that make a denominator zero. These are the restrictions on the variable. b. Keeping the restrictions in mind, solve the equation. 3/(x + 4) - 7 = - 4/(x + 4)

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1
Identify the denominators in the equation: here, the denominators are both \( x + 4 \).
Find the values of \( x \) that make the denominator zero by solving \( x + 4 = 0 \). This gives the restriction \( x \neq -4 \) because division by zero is undefined.
Rewrite the equation \( \frac{3}{x + 4} - 7 = -\frac{4}{x + 4} \) and aim to isolate the variable by eliminating the denominators. Since both fractions have the same denominator, consider multiplying both sides of the equation by \( x + 4 \) to clear the denominators, keeping in mind the restriction \( x \neq -4 \).
After multiplying through by \( x + 4 \), simplify the resulting equation by combining like terms and isolating \( x \) on one side.
Solve the simplified equation for \( x \), then check your solution against the restriction \( x \neq -4 \) to ensure it is valid.

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Restrictions on the Variable

When solving rational equations, identify values that make any denominator zero, as these are undefined and must be excluded from the solution set. For example, if the denominator is (x + 4), then x = -4 is a restriction because it makes the denominator zero.
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Equations with Two Variables

Solving Rational Equations

To solve rational equations, first eliminate denominators by multiplying both sides by the least common denominator (LCD). This transforms the equation into a simpler form without fractions, making it easier to solve for the variable.
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Introduction to Rational Equations

Checking Solutions Against Restrictions

After solving the equation, substitute the solutions back into the original denominators to ensure none violate the restrictions. Any solution that makes a denominator zero must be discarded as extraneous.
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Restrictions on Rational Equations