# SOLUTION: Hello, I have to solve a linear equation but can not find a solution. Please help me how to calculate. x+y+z+u=1 2x+3y-4z+5u=2 3x+4y-3z+6u=0 Thank you sooo much in advanc

Algebra ->  Algebra  -> Coordinate Systems and Linear Equations -> SOLUTION: Hello, I have to solve a linear equation but can not find a solution. Please help me how to calculate. x+y+z+u=1 2x+3y-4z+5u=2 3x+4y-3z+6u=0 Thank you sooo much in advanc      Log On

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 Question 314092: Hello, I have to solve a linear equation but can not find a solution. Please help me how to calculate. x+y+z+u=1 2x+3y-4z+5u=2 3x+4y-3z+6u=0 Thank you sooo much in advance.Answer by Theo(3464)   (Show Source): You can put this solution on YOUR website!It appears there is no solution to this system of equations. The fact becomes clear when you try to eliminate variables and wind up with an equation such as 0 = something that is not equal to 0. Since this equation is not true, the system of equations has no solution. If you had come up with a row that says something like 0 = 0 which is true, then the system of equations would probably have had a dependent solution which means that the solution depends on the value of one of the variables. Since you started with a system of 3 equations in 4 unknowns, the best you could have hoped for would have been a dependent solution. I will go through the solution to show you what happened. ```Your equations are: 1x + 1y + 1z + 1u = 1 2x + 3y - 4z + 5u = 2 3x + 4y - 3z + 6u = 0 Subtract 2 times row 1 from row 2 to get: 0x + 1y - 6z + 3u = 0 Subtract 3 times row 1 from row 3 to get: 0x + 1y - 6z + 3u = -3 You are left with 2 equations in 3 unknowns. They are: 1y - 6z + 3u = 0 1y - 6z + 3u = -3 If you subtract the first equation from the second equation you will get: 0 = -3 Since 0 = -3 is false, this leads to the conclusion that there is no solution to this system of equations. ``` In order for there to be a unique solution to this system of equations, you would have had to have 4 equations in 4 unknowns.