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Solve the following system. y = x + 3 4x + y = 18 (6, 3) (3, 6) (−3, 6) (3, −6)
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(3,6)
\[\begin{pmatrix} 1 & -1 & 3\\ 1 & 4 & 18 \end{pmatrix}\to\begin{pmatrix} 1 & 0 & 6\\ 0 & 1 & 3 \end{pmatrix}\]\(x=3\), \(y=6\)
y-x=3-(1) 4x+y=18-(2) (2)-(1)=>5x=15 x=3 Applying th value of x in eqn (1) y-3=3 y=6
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