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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 13
Capitolo 6, Problema 13

Solve each system in Exercises 5–18. {2x+y=2x+y−z=43x+2y+z=0\(\begin{cases}\) 2x + y = 2 \\ x + y - z = 4 \\ 3x + 2y + z = 0 \(\end{cases}\)

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Write down the system of equations clearly: \[2x + y = 2\] \[x + y - z = 4\] \[3x + 2y + z = 0\]
From the first equation, express \(y\) in terms of \(x\): \[y = 2 - 2x\]
Substitute the expression for \(y\) into the second and third equations to eliminate \(y\): Second equation becomes: \[x + (2 - 2x) - z = 4\] Third equation becomes: \[3x + 2(2 - 2x) + z = 0\]
Simplify both equations to get two equations in terms of \(x\) and \(z\): For the second equation: \[x + 2 - 2x - z = 4\] For the third equation: \[3x + 4 - 4x + z = 0\]
Solve the simplified system of two equations with two variables (\(x\) and \(z\)) using substitution or elimination, then back-substitute to find \(y\).

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