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Solve the system of equations in three variables.
$\left(1,\mathrm{-1},1\right)$
Just as with systems of equations in two variables, we may come across an inconsistent system of equations in three variables, which means that it does not have a solution that satisfies all three equations. The equations could represent three parallel planes, two parallel planes and one intersecting plane, or three planes that intersect the other two but not at the same location. The process of elimination will result in a false statement, such as $\text{\hspace{0.17em}}3=7\text{\hspace{0.17em}}$ or some other contradiction.
Solve the following system.
Looking at the coefficients of $\text{\hspace{0.17em}}x,\text{\hspace{0.17em}}$ we can see that we can eliminate $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ by adding equation (1) to equation (2).
Next, we multiply equation (1) by $\text{\hspace{0.17em}}\mathrm{-5}\text{\hspace{0.17em}}$ and add it to equation (3).
Then, we multiply equation (4) by 2 and add it to equation (5).
The final equation $\text{\hspace{0.17em}}0=2\text{\hspace{0.17em}}$ is a contradiction, so we conclude that the system of equations in inconsistent and, therefore, has no solution.
Solve the system of three equations in three variables.
No solution.
We know from working with systems of equations in two variables that a dependent system of equations has an infinite number of solutions. The same is true for dependent systems of equations in three variables. An infinite number of solutions can result from several situations. The three planes could be the same, so that a solution to one equation will be the solution to the other two equations. All three equations could be different but they intersect on a line, which has infinite solutions. Or two of the equations could be the same and intersect the third on a line.
Find the solution to the given system of three equations in three variables.
First, we can multiply equation (1) by $\text{\hspace{0.17em}}\mathrm{-2}\text{\hspace{0.17em}}$ and add it to equation (2).
We do not need to proceed any further. The result we get is an identity, $\text{\hspace{0.17em}}0=0,$ which tells us that this system has an infinite number of solutions. There are other ways to begin to solve this system, such as multiplying equation (3) by $\text{\hspace{0.17em}}\mathrm{-2},$ and adding it to equation (1). We then perform the same steps as above and find the same result, $\text{\hspace{0.17em}}0=0.$
When a system is dependent, we can find general expressions for the solutions. Adding equations (1) and (3), we have
We then solve the resulting equation for $\text{\hspace{0.17em}}z.$
We back-substitute the expression for $\text{\hspace{0.17em}}z\text{\hspace{0.17em}}$ into one of the equations and solve for $\text{\hspace{0.17em}}y.$
So the general solution is $\text{\hspace{0.17em}}\left(x,\frac{5}{2}x,\frac{3}{2}x\right).\text{\hspace{0.17em}}$ In this solution, $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ can be any real number. The values of $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}z\text{\hspace{0.17em}}$ are dependent on the value selected for $\text{\hspace{0.17em}}x.$
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