Given parameters:
Find if any of the equations are the equivalent;
3a+ 6 = 15
3a = 9
a+ 2= 5
[tex]\frac{1}{3a}[/tex] =1
To find which of the equations are equivalent to one another, let us solve them first. The ones with the same solution are definitely equivalent.
(i) 3a+ 6 = 15
3a = 15 - 6
3a = 9
a = 3
(II) 3a = 9
a = 3
(III) a+ 2= 5
a = 5-2
a = 3
(IV) [tex]\frac{1}{3a}[/tex] =1
1 = 3a
a = [tex]\frac{1}{3}[/tex]
We clearly see that the first three equations are equivalent.
