How do you solve #1/9 - 2/(3b) = 1/18#?

1 Answer
Jul 18, 2017

Answer:

See a solution process below:

Explanation:

Step 1) Subtract #color(red)(1/9)# from each side of the equation to isolate the #b# term while keeping the equation balanced:

#-color(red)(1/9) + 1/9 - 2/(3b) = -color(red)(1/9) + 1/18#

#0 - 2/(3b) = -(2/2 xx color(red)(1/9)) + 1/18#

#-2/(3b) = -2/18 + 1/18#

#-2/(3b) = -1/18#

Step 2) multiply each side of the equation by #color(red)(b)# to move the #b# term to the numerator while keeping the equation balanced:

#color(red)(b) xx -2/(3b) = -1/18 xx color(red)(b)#

#cancel(color(red)(b)) xx -2/(3color(red)(cancel(color(black)(b)))) = (-1)/18 xx color(red)(b)#

#(-2)/3 = (-1)/18b#

Step 3) Multiply each side of the equation by #color(red)(-18)# to solve for #b# while keeping the equation balanced:

#color(red)(-18) xx (-2)/3 = color(red)(-18) xx (-1)/18b#

#-6cancel(color(red)(-18)) xx (-2)/color(red)(cancel(color(black)(3))) = -1cancel(color(red)(-18)) xx (-1)/color(red)(cancel(color(black)(18)))b#

#12 = b#

#b = 12#