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Algebra 2HardSystems of Equations

Systems of Equations · Hard problem

Solve the system for xx, yy and zz.

{4y+2z=14−2x−y+5z=−32−3x+2y−2z=−2\begin{cases} 4 y + 2 z = 14 \\ - 2 x - y + 5 z = -32 \\ - 3 x + 2 y - 2 z = -2 \end{cases}