How do you solve #13- 4z = 40- 7z#?

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Answer 1 · 2017-05-14T04:42:38

See a solution process below:

First, subtract #color(red)(13)# and add #color(blue)(7z)# to each side of the equation to isolate the #z# term while keeping the equation balanced:

#-color(red)(13) + 13 - 4z + color(blue)(7z) = -color(red)(13) + 40 - 7z + color(blue)(7z)#

#0 + (-4 + color(blue)(7))z = 27 + 0#

#3z = 27#

Now, divide each side of the equation by #color(red)(3)# to solve for #z# while keeping the equation balanced:

#(3z)/color(red)(3) = 27/color(red)(3)#

#(color(red)(cancel(color(black)(3)))z)/cancel(color(red)(3)) = 9#

#z = 9#