Question

Solve 2x + y – z = 3
4x – y + 4z = 0
-3y + 2z = 6

(2, -4, -3)

(-2, 4, 3)

(4, 2, -7)

(-4, -2, 7)

Answer 1

Do you know how to do elimination or substitution to solve such a system of equations?

Answer 2

You need to stick around after posting your question if you want help learning how to do this!

Answer 3

By substitution, you might do something like this:

\[2x+y-z=3\]\[4x-y+4z=0\]\[-3y+2z=6\]That first equation has \(y\) with no coefficient, making it easy to solve for \(y\) in terms of \(x, z\):
\[2x+y-z=3\]\[2x-2x+y-z=3-2x\]\[y-z=3-2x\]\[y-z+z=3-2x+z\]\[y=3-2x+z\]Now we take the other two equations and replace \(y\) anywhere we see it with \((3-2x+z)\):

\[4x-(3-2x+z) + 4z = 0\]\[-3(3-2x+z) + 2z = 6\]After you simplify those equations, you get\[6x+3z=3\]\[6x-z=15\]Now you can repeat the process. \(z\) in the second equation is easily isolated, so solve for \(z\) in terms of \(x\). Then substitute that into the \(6x+3z=3\) equation, giving you an equation only in terms of \(x\) which you can easily solve.

Now take the value of \(x\) and plug it into the equation you used for the value of \(z\) (to be clear, this is the equation you got when you solved \(6x-z=15\) for \(z\)). That gives you the value of \(z\). Armed with the values of \(x\) and \(z\), plug them into \(y = 3-2x+z\) to find the value of \(y\).

Now you've found the values of \(x,y,z\) but you aren't quite done. It is important to check that those three values make each of the 3 original equations be true. It is not sufficient to check only one or two of them, as it is possible to make mistakes which will give you a "solution" that only satisfies some of the equations.