A linear equation with three variables represents a plane in three dimensional geometry.A system of linear equations can be solved using the common methods of substitution and elimination.It is not possible to get a graphical solution for a system of three variables on a two dimensional plane.The system of equations in three variables are generally solved using Gaussian row eliminations.

- There is exactly one solution.
- There are infinitely many solutions.
- There is no solution.

Solve the system of Linear Equations

$x-2y+3z=9$ Equation (1)

$-x+3y =-4$ Equation (2)

$2x-5y+5z=17$ Equation (3)

We can eliminate the variable x by manipulating the equations

Adding equations (1) and (2) we get a new equation

$x-2y+3z=9$

$-x+3y =-4$

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$y+3z =5$ Equation (4)

We can multiply the first equation by 2 and add it to equation 3 to get rid of x

$2x-5y+5z=17$ Equation (3)

$-2x+6y = -8$ 2 x Equation (2)

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$y+5z = 9$ Equation (5)

Now equations 4 and 5 are in two variables y and z. We can eliminate the variable y by subtracting equation 4 from 5.

$y+5z = 9$ Equation (5)

$y+3z =5$ Equation (4)

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$2z=4$ Subtraction

$z=2$ Equation solved for the variable z

Now we can substitute

$y+3(2)=5$ Solving this equation we get $y=-1$

Substituting y =-1 in equation (2) we get,

$-x+3(-1)=-4$ Solving this equation we get $x=1$

Hence the solution for the system is

Each of the following operations on a system of linear equations produces an equivalent system of linear equations.

- Interchanging the positions of two equations.
- Multiply one of the equations by a non-zero constant.
- Add a multiple of one of the equations to another equation and replace the later equation.

$x-2y+3z=9$ Equation (1)

$-x+3y =-4$ Equation (2)

$2x-5y+5z=17$ Equation (3)

$x-2y+3z=9$

$ y +3z=5$ Equation (1) is added to equation (2)

$ -y-z=-1$ -2 times equation(1) is added to equation (3).

$x-2y+3z=9$

$ y +3z=5$

$ 2z=4$ Equation (2) in the previous step is added to equation 3.

Now this is an equivalent system of the original system. The equation (3) can be solved easily using division as

The value of

$y+3(2)=5$ → $y=-1$.

Finally substituting back the values

Hence the solution to the given system is x=1,y=-1 and z=2.

- If the resulting equivalent system after doing row operations contains one equation as a true statement without variables, then the system has infinitely many solutions.
- If the resulting equivalent system contains a false statement in an equation, then the system has no solution.