Example Question - variable elimination

System of Linear Equations Problem

<p>The given system of equations is:</p> <p>\[\begin{cases} 3(2x - 3y) - 4(x + y) = 7x - 3y - 5 \\ 7(x - 2y) - 5(2x - y) = -9x + 9y + 7 \end{cases}\]</p> <p>First, expand and simplify both equations:</p> <p>Equation 1: \(6x - 9y - 4x - 4y = 7x - 3y - 5\)</p> <p>Rearrange and combine like terms:</p> <p>\(2x - 13y = 7x - 3y - 5 \Rightarrow -5x + 10y = -5\)</p> <p>Divide by -5 to simplify the equation:</p> <p>\(x - 2y = 1\) ... (i)</p> <p>Equation 2: \(7x - 14y - 10x + 5y = -9x + 9y + 7\)</p> <p>Rearrange and combine like terms:</p> <p>\(-3x - 9y = -9x + 9y + 7 \Rightarrow 6x - 18y = 7\)</p> <p>Divide by 6 to simplify the equation:</p> <p>\(x - 3y = \dfrac{7}{6}\) ... (ii)</p> <p>Now we can solve the system using equations (i) and (ii):</p> <p>From equation (i):</p> <p>\(x = 2y + 1\)</p> <p>Substitute \(x\) from equation (i) into equation (ii):</p> <p>\(2y + 1 - 3y = \dfrac{7}{6}\)</p> <p>\(-y + 1 = \dfrac{7}{6}\)</p> <p>\(-y = \dfrac{7}{6} - 1\)</p> <p>\(-y = \dfrac{1}{6}\)</p> <p>\(y = -\dfrac{1}{6}\)</p> <p>Now substitute \(y\) into equation (i):</p> <p>\(x = 2(-\dfrac{1}{6}) + 1\)</p> <p>\(x = -\dfrac{1}{3} + 1\)</p> <p>\(x = \dfrac{2}{3}\)</p> <p>The solution to the system of equations is \(x = \dfrac{2}{3}\), \(y = -\dfrac{1}{6}\).</p>