\( \left. \begin{array} { | l | l } \hline 1.3 x - 6 y = - 12 \\ { y = \frac { 1 } { 2 } x + 2 } \\ { 6 y = 2 y + 2 } & { 2 } \\ \hline \end{array} \right. \)

It looks like you have a system of equations and possibly a table format. Let's break down the equations you provided: 1. \( 1.3x - 6y = -12 \) 2. \( y = \frac{1}{2}x + 2 \) 3. \( 6y = 2y + 2 \) Let's solve these equations step by step. ### Step 1: Solve the third equation The third equation is: \[ 6y = 2y + 2 \] Subtract \(2y\) from both sides: \[ 6y - 2y = 2 \] This simplifies to: \[ 4y = 2 \] Now, divide both sides by 4: \[ y = \frac{2}{4} = \frac{1}{2} \] ### Step 2: Substitute \(y\) into the second equation Now that we have \(y = \frac{1}{2}\), we can substitute this value into the second equation to find \(x\): \[ \frac{1}{2} = \frac{1}{2}x + 2 \] Subtract 2 from both sides: \[ \frac{1}{2} - 2 = \frac{1}{2}x \] Convert 2 to a fraction with a denominator of 2: \[ \frac{1}{2} - \frac{4}{2} = \frac{1}{2}x \] This simplifies to: \[ -\frac{3}{2} = \frac{1}{2}x \] Now, multiply both sides by 2 to solve for \(x\): \[ -3 = x \] ### Step 3: Verify with the first equation Now we have \(x = -3\) and \(y = \frac{1}{2}\). Let's verify these values in the first equation: \[ 1.3(-3) - 6\left(\frac{1}{2}\right) = -12 \] Calculating the left side: \[ -3.9 - 3 = -12 \] This is correct, as \(-3.9 - 3 = -12\). ### Conclusion The solution to the system of equations is: \[ x = -3, \quad y = \frac{1}{2} \]