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Ch. 5 - Systems of Equations and Inequalities
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
All textbooksBlitzer 8th EditionCh. 5 - Systems of Equations and InequalitiesProblem 5
Chapter 6, Problem 5

In Exercises 1–18, solve each system by the substitution method. {y=x2−4x−10y=−x2−2x+14\(\begin{cases}\)y = x^2 - 4x - 10 \(\y\) = -x^2 - 2x + 14\(\end{cases}\)

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Start with the given system of equations: \(y = x^{2} - 4x - 10\) and \(y = -x^{2} - 2x + 14\).
Since both expressions are equal to \(y\), set them equal to each other to find \(x\): \(x^{2} - 4x - 10 = -x^{2} - 2x + 14\).
Combine like terms by bringing all terms to one side to form a quadratic equation: \(x^{2} - 4x - 10 + x^{2} + 2x - 14 = 0\).
Simplify the equation to standard quadratic form: \(2x^{2} - 2x - 24 = 0\).
Solve the quadratic equation for \(x\) using factoring, completing the square, or the quadratic formula, then substitute each \(x\) value back into one of the original equations to find the corresponding \(y\) values.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

System of Equations

A system of equations consists of two or more equations with the same variables. The goal is to find values for the variables that satisfy all equations simultaneously. In this problem, the system involves two equations with variables x and y.
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Substitution Method

The substitution method involves solving one equation for one variable and then substituting that expression into the other equation. This reduces the system to a single equation with one variable, making it easier to solve.
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Quadratic Equations

Quadratic equations are polynomial equations of degree two, typically in the form ax² + bx + c = 0. Solving quadratic equations often involves factoring, completing the square, or using the quadratic formula. Here, both equations are quadratic in x, which affects the substitution and solution process.
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Related Practice
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