1. bc/3 + a/xy + 3/ab
= b^2cxya + 3a^2b + 9xy/3xyab
2. (5/3y) + 2y - 12
=11/3y - 2
3. ?
Hope this helps and right! :)) Good luck
All you do is subtract the two numbers
Start with

Multiply the whole equation by 2. Since 2 is positive, we don't need to switch the inequality sign:

Subtract 3 from both sides:

Answer:

Step-by-step explanation:

Start by factoring out a 5:

We need to find two integers that have a product of 12, and a sum of -7:
(-3)(-4)=12
-3-4=-7
We can split -7x into -3x and -4x

Factor each half separately:
![5[x(x-3)-4(x-3)]](https://tex.z-dn.net/?f=5%5Bx%28x-3%29-4%28x-3%29%5D)
Since x and -4 are both being multiplied by x-3, we can combine them:
