Show that this equation is NOT consipten [ 2 x+6 y+11=0 6 x+20 y-6 z-3=0 6 y-18 z+1=0 ]

To determine if the given system of equations is consistent, we need to check if there exists a set of values for \(x\), \(y\), and \(z\) that satisfies all three equations simultaneously. If such a set exists, the system is consistent; otherwise, it is not.<br /><br />Let's analyze each equation:<br /><br />1. \(2x + 6y + 11 = 0\)<br />2. \(6x + 20y - 6z - 3 = 0\)<br />3. \(6y - 18z + 1 = 0\)<br /><br />First, let's solve the third equation for \(z\):<br /><br />\[6y - 18z + 1 = 0\]<br />\[18z = 6y + 1\]<br />\[z = \frac{6y + 1}{18}\]<br />\[z = \frac{y + 1}{3}\]<br /><br />Now, substitute \(z = \frac{y + 1}{3}\) into the second equation:<br /><br />\[6x + 20y - 6\left(\frac{y + 1}{3}\right) - 3 = 0\]<br />\[6x + 20y - 2(y + 1) - 3 = 0\]<br />\[6x + 20y - 2y - 2 - 3 = 0\]<br />\[6x + 18y - 5 = 0\]<br /><br />Now, we have two equations with two variables (\(x\) and \(y\)):<br /><br />1. \(2x + 6y + 11 = 0\)<br />2. \(6x + 18y - 5 = 0\)<br /><br />Let's solve these two equations simultaneously. We can use the elimination method. Multiply the first equation by 3:<br /><br />\[3(2x + 6y + 11) = 0\]<br />\[6x + 18y + 33 = 0\]<br /><br />Now, subtract the second equation from this result:<br /><br />\[(6x + 18y + 33) - (6x + 18y - 5) = 0\]<br />\[6x + 18y + 33 - 6x - 18y + 5 = 0\]<br />\[38 = 0\]<br /><br />This is a contradiction, which means there is no set of values for \(x\), \(y\), and \(z\) that satisfies all three equations simultaneously. Therefore, the given system of equations is not consistent.