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

Question

Eli Assouline · Accepted Answer

#y=-9#

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#

Eva Abramson · Answer

#y=-9#

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!

Isaiah Anthony · Answer

undefined To solve for y in the equation (y+5)/2 - y/3 = 1, we can simplify the equation by finding a common denominator and combining like terms. Multiplying the first term by 3/3 and the second term by 2/2, we get (3(y+5))/6 - (2y)/6 = 1. Simplifying further, we have (3y + 15 - 2y)/6 = 1. Combining like terms, we get (y + 15)/6 = 1. Multiplying both sides of the equation by 6, we have y + 15 = 6. Subtracting 15 from both sides, we get y = -9. Therefore, the solution to the equation is y = -9.