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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 7

Solve each system in Exercises 5–18. {4x−y+2z=11x+2y−z=−12x+2y−3z=−1\(\begin{cases}\) 4x - y + 2z = 11 \\ x + 2y - z = -1 \\ 2x + 2y - 3z = -1 \(\end{cases}\)

Guida verificata passo dopo passo
1
Write down the system of equations clearly: \[4x - 0y + 2z = 11\] \[x + 2y - z = -1\] \[2x + 2y - 3z = -1\]
Since the first equation has no y term, focus on expressing one variable in terms of the others from one of the simpler equations. For example, from the second equation, solve for \(x\) in terms of \(y\) and \(z\): \[x = -1 - 2y + z\]
Substitute the expression for \(x\) from step 2 into the first and third equations to eliminate \(x\). This will give you two equations with only \(y\) and \(z\) as variables.
Solve the resulting two-variable system from step 3 using either substitution or elimination to find values for \(y\) and \(z\).
Once you have \(y\) and \(z\), substitute these values back into the expression for \(x\) from step 2 to find the value of \(x\). This completes the solution for the system.

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