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**topspin1617** In a nutshell, you are exactly right. From the first two equations alone, we can see $\displaystyle z=1$. Plugging this in to the equations, we just get the same equation repeated three times: $\displaystyle w+x+y=4$. One equation in three values means that two of the variables are free (can be anything), and the value of the third depends on the values of the first two. So if, for example, $\displaystyle x=s,y=t$, then the set of all solutions would look like vectors (or points) of the form

$\displaystyle \left(\begin{array}{c}{x\\y\\z\\w}\end{array}\righ t)=\left(\begin{array}{c}{4-s-t\\s\\t\\1}\end{array}\right)=\left(\begin{array}{ c}{4\\0\\0\\1}\end{array}\right)+s\left(\begin{arr ay}{c}{-1\\1\\0\\0}\end{array}\right)+t\left(\begin{array} {c}{-1\\0\\1\\0}\end{array}\right)$

(many linear algebra tests would write the solution in the latter form, separating pieces corresponding to the independent parameters. But you don't HAVE to; there is absolutely nothing wrong with writing it the other way).