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Solve a System of Linear Equations with Three Variables:

x-2y+3z=1x+y-3z=73x-4y+5z=7

Short Answer

Expert verified

Solution of the system of equations is(5+z,2+2z,z).

Step by step solution

01

Step 1. Given Information.

We have been given system of equations :

x-2y+3z=1x+y-3z=73x-4y+5z=7

02

Step 2. Using Elimination method to solve system of equations.

Multiply -1in second equation to get opposite coefficients of x

x+y-3z=7

-x-y+3z=-7

03

Step 3. Add the equations to get two variable equation.

Add the equations : x-2y+3z=1-x-y+3z=-7

-3y+6z=-6

-y+2z=-2

04

Step 4. Multiply -3 in first equation, to get opposite coefficients of x.

x-2y+3z=1

-3(x-2y+3z)=-3(1)

-3x+6y-9z=-3

Now, add the equations : -3x+6y-9z=-33x-4y+5z=7

2y-4z=4

y-2z=2

05

Step 5. Add the equations.

-y+2z=-2y-2z=2

0=0

This is True.

System has infinitely many solutions.

06

Step 6. Represent the solution showing how x,y are dependent on z.

So, we have y-2z=2

y=2z+2

And, equation x-2y+3z=1

Substitute -

x-2(2z+2)+3z=1

x-4z-4+3z=1

x-z=5

x=5+z

Solution of the system of equations is(5+z,2+2z,z).

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