Consider the following system of linear equations. 2x-6y=-14 8x-22y=-52 Solve the system by completing the steps below to produce a reduced row-echelon form. R₁ and R₂ denote the first and second rows, respectively. The arrow notation (→) means the expression/matrix on the left becomes the expression/matrix on the right once the row operations are performed. (a) For each step below, enter the coefficient for the row operation. The first matrix in Step 1 is the augmented matrix for the given system of equations. Step 1: 2 -6 - 14 8-22 i -52 Step 2: -7 8 22 -52 Step 3: -3 ≈ 1-3 -7 0 2 I 4 I -3 -7 1 0 1 x = 2 (b) Give the solution. 0 R₁ → R₁ R₁ + R₂ R₂ Step 4: Enter the coefficient for the row operations and the missing entries in the resulting matrix. R₂ R₂ y = 0 R₂ + R₁ R₁ 2 8 - -3 -7 -22 i -52 1-3 ! [213] 4 -3 [213] 0 10 18 0 X

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