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Solve the system of equations:

2x−2y+3z=64x−3y+2z=0−2x+3y−7z=1

Short Answer

Expert verified

There is no solution of given system of linear equations.

Step by step solution

01

Step 1. Given Information

We are given a system of linear equation,

2x−2y+3z=6-(1)4x−3y+2z=0-(2)−2x+3y−7z=1-(3)

02

Step 2. Solving the equations 

Adding first and third equation, we get

2x-2x-2y+3y+3z-7z=6+1y-4z=7-(4)

Multiplying by 2in first equation and subtraction it from second equation,

2(2x-2y+3z)=2(6)4x-4y+12z=12

Subtracting, we get

role="math" localid="1644577982732" 4x-4x-4y+3y+6z-2z=12-0-y+4z=12-(5)

Now, adding fourth and fifth equation, we get

y-y-4z+4z=7+120≠19

This is not true hence the given system has no solution.

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