How do you evaluate #z-w+y# if w=6, x=4, y=-2, z=6?

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Answer 1 · 2017-06-06T01:49:38

See a solution process below:

The #x# valuable is not in the expression and is therefore extraneous and does not need to be considered in the solution.

We can then substitute:

#color(red)(6)# for #color(red)(w)#

#color(blue)(-2)# for #color(blue)(y)#

#color(green)(6)# for #color(green)(z)#

in the expression and calculate the result:

#color(green)(z) - color(red)(w) + color(blue)(y)# becomes:

#color(green)(6) - color(red)(6) + color(blue)(-2) =>#

#0 + color(blue)(-2) =>#

#-2#