How do you solve #19a - 3(a - 6) = 66# for #a#?

First expand the term in parenthesis by multiplying each term within the parenthesis by the term outside the parenthesis:

#19a - color(red)(3)(a - 6) = 66#

#19a - (color(red)(3) * a) + (color(red)(3) * 6) = 66#

#19a - 3a + 18 = 66#

Next, combine like terms:

#(19 - 3)a + 18 = 66#

#16a + 18 = 66#

Then, subtract #color(red)(18)# from each side of the equation to isolate the #a# term while keeping the equation balanced:

#16a + 18 - color(red)(18) = 66 - color(red)(18)#

#16a + 0 = 48#

#16a = 48#

Now, divide each side of the equation by #color(red)(16)# to solve for #a# while keeping the equation balanced:

#(16a)/color(red)(16) = 48/color(red)(16)#

#(color(red)(cancel(color(black)(16)))a)/cancel(color(red)(16)) = 3#

#a = 3#

#19a - 3(a-6) = 66#

#19a - 3a + 18 = 66#

#19a - 3a color(red)(cancel(color(black)(+ 18) - 18)) = 66 color(red)(- 18)#

#19a - 3a = 66 - 18#

#16a = 66 - 18#

#16a = 48#

#color(red)(cancel(color(black)(16))) a color(red)(cancel(÷ 16)) = 48 ÷ 16#

#a = 48 ÷ 16#

#color(blue)(a = 3#

We can substitute #a# for #3# to prove our answer.

#19 xx 3 - 3(3 - 6) = 66#

#19 xx 3 - 3 xx -3 = 66#

#57 - 3 xx -3 = 66#

#57 - -9 = 66#

#57 + 9 = 66#