91Ó°ÊÓ

We will rewrite the given system of equations as a matrix representation: $$ \left\{\begin{array}{l} 2x - y = -1\\ 4x - 7y = 3 \end{array}\right. $$ The matrix form is an augmented matrix: $$ \left[ {\begin{array}{cc|c} 2 & -1 & -1 \\ 4 & -7 & 3 \end{array} } \right] $$

Our goal is to get ones on the diagonal and zeros below it. Start by multiplying row 1 by \(\frac{1}{2}\), to get a one in the top left corner: $$ \left[ {\begin{array}{cc|c} 1 & -\frac{1}{2} & -\frac{1}{2} \\ 4 & -7 & 3 \end{array} } \right] $$ Now, we need to eliminate the 4 in row 2. To do this, subtract 4 times row 1 from row 2: $$ \left[ {\begin{array}{cc|c} 1 & -\frac{1}{2} & -\frac{1}{2} \\ 0 & -6 & 5 \end{array} } \right] $$ Next, divide row 2 by -6 to get a 1 in the bottom right corner: $$ \left[ {\begin{array}{cc|c} 1 & -\frac{1}{2} & -\frac{1}{2} \\ 0 & 1 & -\frac{5}{6} \end{array} } \right] $$ Finally, add \(\frac{1}{2}\) times row 2 to row 1 to eliminate the -\(\frac{1}{2}\) in row 1: $$ \left[ {\begin{array}{cc|c} 1 & 0 & -\frac{1}{3} \\ 0 & 1 & -\frac{5}{6} \end{array} } \right] $$ Now that the matrix is in row echelon form, we can read the solution directly.

The matrix in row echelon form represents the system: $$ \left\{\begin{array}{l} x + 0y = -\frac{1}{3}\\ 0x + y = -\frac{5}{6} \end{array}\right. $$ Which we can simplify to: $$ \left\{\begin{array}{l} x = -\frac{1}{3}\\ y = -\frac{5}{6} \end{array}\right. $$ Now, we can plug these values back into the original system of equations to check that they are a valid solution: Equation 1: \( 2x + 5 = 4 + y\) $$ 2 \left(-\frac{1}{3}\right) + 5 = 4 -\frac{5}{6} $$ $$ \frac{5}{3} = \frac{13}{6} $$ Equation 2: \(7y = 4x - 3\) $$ 7 \left(-\frac{5}{6}\right) = 4 \left(-\frac{1}{3}\right) - 3 $$ $$ \frac{-35}{6} = \frac{-15}{3} $$ The values of x and y make both equations true, so the solution to the system is: $$ \boxed{x = -\frac{1}{3},\ y = -\frac{5}{6}} $$