We start with the given system of equations: 3x−2y=3(1)−4x+y=6(2) \begin{array}{l} 3x - 2y = 3 \quad (1) \\ -4x + y = 6 \quad (2) \end{array} 3x−2y=3(1)−4x+y=6(2)
We can represent the system in augmented matrix form [A∣b] [A | b] [A∣b]: [A∣b]=[3−23−416] \left[ A | b \right] = \left[ \begin{array}{ccc} 3 & -2 & 3 \\ -4 & 1 & 6 \\ \end{array} \right] [A∣b]=[3−4−2136]
We perform row operations to reduce the matrix to row echelon form. The steps are as follows:
Divide the first row by 3: [A∣b]=[1−231−416] \left[ A | b \right] = \left[ \begin{array}{ccc} 1 & -\frac{2}{3} & 1 \\ -4 & 1 & 6 \\ \end{array} \right] [A∣b]=[1−4−32116]
Add 4 times the first row to the second row: [A∣b]=[1−2310−5310] \left[ A | b \right] = \left[ \begin{array}{ccc} 1 & -\frac{2}{3} & 1 \\ 0 & -\frac{5}{3} & 10 \\ \end{array} \right] [A∣b]=[10−32−35110]
Multiply the second row by −35-\frac{3}{5}−53: [A∣b]=[1−23101−6] \left[ A | b \right] = \left[ \begin{array}{ccc} 1 & -\frac{2}{3} & 1 \\ 0 & 1 & -6 \\ \end{array} \right] [A∣b]=[10−3211−6]
Add 23\frac{2}{3}32 times the second row to the first row: [A∣b]=[10−301−6] \left[ A | b \right] = \left[ \begin{array}{ccc} 1 & 0 & -3 \\ 0 & 1 & -6 \\ \end{array} \right] [A∣b]=[1001−3−6]
From the final row echelon form, we can read the solutions: x=−3y=−6 x = -3 \\ y = -6 x=−3y=−6
The solution to the system of equations is: (x,y)=(−3,−6) \boxed{(x, y) = (-3, -6)} (x,y)=(−3,−6)
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