Some information was missing so I am attaching the complete question.
Given Equation is:
[tex]\frac{3y}{5}\;+\;\frac{3}{2}\;=\;\frac{7y}{10}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;------\;\;(1)[/tex]
Step (a)
Let, A = 5 * 3 * 10
A = 150
Step (b)
Multiplying both sides of equation (1) with A = 150, we get,
[tex]90y_{_1}\; + \;225\; =\; 105y_{_1}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;------\;\;(a)[/tex]
Multiplying both sides of equation (1) by 10 we get,
[tex]6y_{_2}\; + \;15\; =\; 7y_{_2}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;------\;\;(b)[/tex]
Step (c)
Solving Equation (a), we get
[tex]105y_{_1}\;-\; 90y_{_1}\;=\;225\\\\15y_{_1}\;= \;225\\\\y_{_1}\;=\;\frac{225}{15}\\\\y_{_1}\;=\;15[/tex]
Solving Equation (b), we get
[tex]7y_{_{2}}\; -\; 6y_{_{2}}\; = \;15\\\\y_{_{2}}\;=\; 15[/tex]
Step (d)
[tex]y{_{_1}} \;= \;15; \;\;\;\;\;\&\;\;\;\;\;y{_{_2}}\; = \;15;\\\\\therefore y{_{_1}}\; =\; y{_{_2}}[/tex]
Step (e)
Both methods work the same because as long as we multiply (or divide) the same amount to the total equation, the equation balances itself out. For example, consider an equation
x = 2;
If we multiply it by 10 on both sides we will get
10x = 20,
If we simplify the above equation, however, we will get the same initial equation.
x = 20/10
x = 2
