How do you solve #1/2z+1/3=-2/5#?

2 Answers
Apr 23, 2017

Answer:

See the entire solution process below:

Explanation:

First, multiply each side of the equation by #color(red)(30)# to eliminate the fractions while keeping the equation balanced. #color(red)(3)# is the Lowest Common Denominator of the three fractions:

#color(red)(30)(1/2z + 1/3) = color(red)(30) * -2/5#

#(color(red)(30) * 1/2z) + (color(red)(30) * 1/3) = cancel(color(red)(30))6 * -2/color(red)(cancel(color(black)(5)))#

#(cancel(color(red)(30)) 15 * 1/color(red)(cancel(color(black)(2)))z) + (cancel(color(red)(30)) 10 * 1/color(red)(cancel(color(black)(3)))) = 6 * -2#

#(15 * 1z) + (10 * 1) = -12#

#15z + 10 = -12#

Next, subtract #color(red)(10)# from each side of the equation to isolate the #z# term while keeping the equation balanced:

#15z + 10 - color(red)(10) = -12 - color(red)(10)#

#15z + 0 = -22#

#15z = -22#

Now, divide each side of the equation by #color(red)(15)# to solve for #z# while keeping the equation balanced:

#(15z)/color(red)(15) = -22/color(red)(15)#

#(color(red)(cancel(color(black)(15)))z)/cancel(color(red)(15)) = -22/15#

#z = -22/15#

Apr 23, 2017

Answer:

#z=-22/15#

Explanation:

First, let's multiply all terms by 30 to remove the fractions.
Doing this, we get:
#15z+10=-12#

Then, we simplify.
#15z=-22#

#z=-22/15#