Find all values of x such that y = 0. y = 1/(5x + 5) - 3/(x + 1) + 7/5
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- 0. Review of Algebra4h 18m
- 1. Equations & Inequalities3h 18m
- 2. Graphs of Equations1h 43m
- 3. Functions2h 17m
- 4. Polynomial Functions1h 44m
- 5. Rational Functions1h 23m
- 6. Exponential & Logarithmic Functions2h 28m
- 7. Systems of Equations & Matrices4h 5m
- 8. Conic Sections2h 23m
- 9. Sequences, Series, & Induction1h 22m
- 10. Combinatorics & Probability1h 45m
1. Equations & Inequalities
Linear Equations
Problem 87a
Textbook Question
The equations in Exercises 79–90 combine the types of equations we have discussed in this section. Solve each equation. Then state whether the equation is an identity, a conditional equation, or an inconsistent equation. 4/(x - 2) + 3/(x + 5) = 7/(x + 5)(x - 2)
Verified step by step guidance1
Rewrite the equation with a common denominator. The least common denominator (LCD) for the terms is \((x - 2)(x + 5)\). Multiply through by the LCD to eliminate the fractions.
After multiplying through by \((x - 2)(x + 5)\), simplify each term. The first term becomes \(4(x + 5)\), the second term becomes \(3(x - 2)\), and the right-hand side becomes \(7\). This results in the equation \(4(x + 5) + 3(x - 2) = 7\).
Distribute the constants in the parentheses. Expand \(4(x + 5)\) to \(4x + 20\) and \(3(x - 2)\) to \(3x - 6\). Combine these terms to form \(4x + 20 + 3x - 6 = 7\).
Combine like terms on the left-hand side. Combine \(4x\) and \(3x\) to get \(7x\), and combine \(20\) and \(-6\) to get \(14\). The equation simplifies to \(7x + 14 = 7\).
Isolate \(x\) by subtracting \(14\) from both sides, resulting in \(7x = -7\). Then divide both sides by \(7\) to solve for \(x\). Finally, check the solution in the original equation to determine if it is valid, and classify the equation as an identity, conditional, or inconsistent.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Types of Equations
In algebra, equations can be classified into three main types: identities, conditional equations, and inconsistent equations. An identity holds true for all values of the variable, a conditional equation is true for specific values, and an inconsistent equation has no solutions. Understanding these classifications is essential for determining the nature of the given equation.
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Solving Rational Equations
Rational equations involve fractions with polynomials in the numerator and denominator. To solve these equations, one typically finds a common denominator to eliminate the fractions, allowing for easier manipulation and simplification. This process is crucial for isolating the variable and finding potential solutions.
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Checking Solutions
After solving an equation, it is important to check the solutions by substituting them back into the original equation. This verification process ensures that the solutions are valid and helps identify any extraneous solutions that may arise, particularly in rational equations where division by zero can occur.
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