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Ch. 5 - Systems of Equations and Inequalities
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBlitzer 8th EditionCh. 5 - Systems of Equations and InequalitiesProblema 39
Capitolo 6, Problema 39

In Exercises 29–42, solve each system by the method of your choice. {y=(x+3)2x+2y=−2\(\begin{cases}\) y = (x + 3)^2 \\ x + 2y = -2 \(\end{cases}\)

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Start with the given system of equations: \(y = (x+3)^2\) and \(x + 2y = -2\).
Since \(y\) is already expressed in terms of \(x\) in the first equation, substitute \(y = (x+3)^2\) into the second equation to eliminate \(y\).
After substitution, the second equation becomes \(x + 2(x+3)^2 = -2\). Expand the squared term \((x+3)^2\) to get \(x + 2(x^2 + 6x + 9) = -2\).
Distribute the 2 across the terms inside the parentheses: \(x + 2x^2 + 12x + 18 = -2\).
Combine like terms and rearrange the equation to standard quadratic form: \(2x^2 + 13x + 18 + 2 = 0\), which simplifies to \(2x^2 + 13x + 20 = 0\). Then solve this quadratic equation for \(x\) using factoring, completing the square, or the quadratic formula.

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