Question

Linear equations in three variablesI understand linear equations in three variables, but in this problem they took the y variable out of the first linear equation. What changes? How do I do it?

Answer 1

`\begin{gathered} x\text{ - z = -5 ---------(1)} \\ 2x\text{ + 3y + 2z = 12 -------(2)} \\ x\text{ + y + 3z = 13 --------(3)} \end{gathered}`

To eliminate y, we multiply equation (3) by 3;

`3x\text{ + 3y + 9z = 39 ------------(4)}`

Equation (4) - Equation (2);

`\begin{gathered} 3x-2x\text{ + 3y-3y + 9z - 2z = 39 - 12} \\ x\text{ + 7z = 27 --------------------(5)} \end{gathered}`

Solving equations (1) and (5) simultaneously;

`\begin{gathered} x-z\text{ = -5 ---------(1)} \\ x+7z\text{ = 27 ---------(5)} \\ (5)\text{ - (1)} \\ x-x\text{ + 7z-(-z) = 27-(-5)} \\ 7z+z\text{ = 27+5} \\ 8z\text{ = 32} \\ (8z)/(8)=\text{ }(32)/(8) \\ z\text{ = 4} \end{gathered}`

Substituting 4 for z in (1);

`\begin{gathered} x-z=-5 \\ x\text{ - 4 = -5} \\ x\text{ = -5+4} \\ x=-1 \end{gathered}`

Since x = -1 and z = 4, we can now find y. Making use of the equation (3);

`\begin{gathered} x+y+3z=13 \\ -1+y\text{ +3(4) = 13} \\ -1+y+12\text{ = 13} \\ y\text{ +11 = 13} \\ y=13-11 \\ y=2 \end{gathered}`

CONCLUSION:

`\begin{gathered} x=-1 \\ y=2 \\ z=4 \end{gathered}`