How do you solve #3-4(2k-5)=71# using the distributive property?

2 Answers
Mar 12, 2017

Answer:

See the entire solution process below:

Explanation:

First, expand the term in parenthesis on the left side of the equation by multiplying each term within the parenthesis by #color(red)(4)# - the term outside the parenthesis:

#3 - color(red)(4)(2k - 5) = 71#

#3 - (color(red)(4)xx2k) + (color(red)(4)xx5) = 71#

#3 - 8k + 20 = 71#

#-8k + 20 + 3 = 71#

#-8k + 23 = 71#

Next, subtract #color(red)(23)# from each side of the equation to isolate the #k# term while keeping the equation balanced:

#-8k + 23 - color(red)(23) = 71 - color(red)(23)#

#-8k + 0 = 48#

#-8k = 48#

Now, divide each side of the equation by #color(red)(-8)# to solve for #k# while keeping the equation balanced:

#(-8k)/color(red)(-8) = 48/color(red)(-8)#

#(color(red)(cancel(color(black)(-8)))k)/cancel(color(red)(-8)) = -6#

#k = -6#

Mar 12, 2017

Answer:

#k=-6#

Explanation:

distribute the bracket to begin with.

#3-8k+20=71#

#rArr-8k+23=71#

subtract 23 from both sides.

#-8kcancel(+23)cancel(-23)=71-23#

#rArr-8k=48#

divide both sides by - 8

#(cancel(-8) k)/cancel(-8)=48/(-8)#

#rArrk=-6#

#color(blue)"As a check"#

Substitute this value into the left side of the equation and if equal to the right side then it is the solution.

#"left side "=3-4((2xx-6)-5)#

#color(white)(left side)=3-(4xx(-17))#

#color(white)(left side)=3-(-68)=3+68=71#

#rArrk=-6" is the solution"#