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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 5
Capitolo 6, Problema 5

Solve each system in Exercises 5–18.
{x+y+2z=11x+y+3z=14x+2y−z=5\(\begin{cases}\) x + y + 2z = 11 \\ x + y + 3z = 14 \\ x + 2y - z = 5 \(\end{cases}\)

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1
Write down the system of equations clearly: \(\begin{cases} x + 0y + 2z = 11 \\ x + 0y + 3z = 14 \\ x + 2y - 0z = 5 \end{cases}\)
Notice that the first two equations both have \(x\) and \(z\) terms but no \(y\). Use these two equations to eliminate \(x\) or \(z\) by subtracting one equation from the other.
Subtract the first equation from the second: \( (x + 0y + 3z) - (x + 0y + 2z) = 14 - 11 \) which simplifies to an equation involving only \(z\).
Solve the resulting equation for \(z\). Once you have \(z\), substitute this value back into one of the first two equations to solve for \(x\).
With \(x\) and \(z\) known, substitute both into the third equation \(x + 2y = 5\) to solve for \(y\).

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