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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
Capitolo 6, Problema 41

In Exercises 29–42, solve each system by the method of your choice. {x2+y2+3y=222x+y=−1\(\begin{cases}\) x^2 + y^2 + 3y = 22 \\ 2x + y = -1 \(\end{cases}\)

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Start by examining the given system of equations: \(x^2 + y^2 + 3y = 22\) and \(2x + y = -1\).
From the linear equation \(2x + y = -1\), solve for \(y\) in terms of \(x\): \(y = -1 - 2x\).
Substitute the expression for \(y\) into the first equation to eliminate \(y\): \(x^2 + (-1 - 2x)^2 + 3(-1 - 2x) = 22\).
Expand and simplify the resulting equation to form a quadratic equation in terms of \(x\) only.
Solve the quadratic equation for \(x\), then substitute each solution back into \(y = -1 - 2x\) to find the corresponding \(y\) values.

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