How do you solve for y in #(y+5)/ 2 - y/3 =1#?

2 Answers
Jun 10, 2018

Answer:

#y=-9#

Explanation:

Solve:

#(y+5)/2-y/3=1#

The LCM of #2# and #3# is #6#. Multiply both sides by #6#.

#(6(y+5))/2-(6(y))/3=1(6)#

Simplify.

#(color(red)cancel(color(black)(6))^3(y+5))/color(red)cancel(color(black)(2))^1-(color(red)cancel(color(black)(6))^2(y))/color(red)cancel(color(black)(3))^1=1(6)#

#3(y+5)-2y=6#

Expand.

#3y+15-2y=6#

Subtract #15# from both sides.

#3y-2y=6-15#

Simplify.

#y=-9#

Jun 10, 2018

Answer:

#y=-9#

Explanation:

As we know, when subtracting any fraction, we must have like denominators. To achieve this, let's multiply the first one by #3/3# and the second by #2/2#. We now have

#color(blue)((3/3))(y+5)/2-color(blue)((2/2))(y/3)=1#

Which simplifies to

#(3(y+5)-2y)/6=1#

Distributing the #3#, we now have

#(3y+15-2y)/6=1#

Combining like terms in the numerator, we get

#(y+15)/6=1#

Multiply both sides by #6# to get

#y+15=6#

Lastly, subtracting #15# from both sides gives us

#y=-9#

Hope this helps!