How do you solve #7(5a-4)-1=14-8a#?

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Answer 1 · 2017-02-27T15:00:15

See the entire solution process below:

First, expand the term in parenthesis on the left side of the equation by multiplying each term within the parenthesis by #color(red)(7)#:

#(color(red)(7) xx 5a) - (color(red)(7) xx 4) - 1 = 14 - 8a#

#35a - 28 - 1 = 14 - 8a#

#35a - 29 = 14 - 8a#

Next, add #color(red)(29)# and #color(blue)(8a)# to each side of the equation to isolate the #a# term while keeping the equation balanced:

#35a - 29 + color(red)(29) + color(blue)(8a) = 14 - 8a + color(red)(29) + color(blue)(8a)#

#35a + color(blue)(8a) - 29 + color(red)(29) = 14 + color(red)(29) - 8a + color(blue)(8a)#

#43a - 0 = 43 - 0#

#43a = 43#

Now, divide each side of the equation by #color(red)(43)# to solve for #a# while keeping the equation balanced:

#(43a)/color(red)(43) = 43/color(red)(43)#

#(color(red)(cancel(color(black)(43)))a)/cancel(color(red)(43)) = 1#

#a = 1#