Algebra help - systems of equations?

Question

Here's the problem:

Solve - Show work and justify your answer

4x - 2y + 6z = 12
2x - y + 3y = 5
3x + 2y + z = 6

I can use any method I want to get the answer, and have been trying to using the elimination method. And on the second line, it says 3y, and was wondering if this could have bee a typo. I haven't been able get a working answer. Can this even be solved? I'd appreciate any help. Thanks.

Answer 1

This system has no solution with or without the typo. For the system to have a solution then one point (x, y, z) would have to work in all three equations.

If the 3y is really a 3z, then the first two equations represent parallel lines. (Divide the first equation by 2 and notice that the coefficients for the first two equations will match)

Even if you use the 3y 'as is' .. and if you attempt to solve the system by elimination, you will always end up with two equations so that the coefficients of the x's and the z's are congruent. This means that if you eliminate one variable, you end up eliminating both of them creating a false statement.

Therefore, this system has no solution. (Any pairing of the three lines are parallel or skew)

Answer 2

I would assume the 3y is a typo, if not it will make the second equation easier to solve. If you know matrices you can solve that way, if not elimination is probably your best bet.

Answer 3

I'm guessing that it's a typo, too. But if not, then your second equation is simply
2x -2y +0z = 5

Use Gauss-Jordan to solve

Answer 4

divide first line by 2
2x -y +3z =6
compare second (yes assume misprint)
2x -y +3z = 5

the LHS is the same (so you dont have enough info to find "a" solution) and the RHS is different ... so the equations are inconsistent and there are no solutions

Answer 5

the only hope you got is to combine like terms... sorryyy