How do you evaluate #y-:2+x# using #x=1# and #y=2#?

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Answer 1 · 2017-06-07T19:27:53

See a solution process below:

First, substitute #color(red)(1)# for #color(red)(x)# and substitute #color(blue)(2)# for #color(blue)(y)#:

#color(blue)(y) -: 2 + color(red)(x)# becomes:

#color(blue)(2) -: 2 + color(red)(1)#

Next execute the division operation first:

#(color(blue)(2) -: 2) + color(red)(1) => 1 + color(red)(1)#

Now, execute the addition operation:

#1 + color(red)(1) => 2#