transforming formulas

we isolate the variable by removing the constant (+2) by subtracting 2

y-2 = 15x + 2 - 2

which leaves us with y - 2= 15x

next step we isolate by dividing both sides by the coefficient (15)

 $\frac{\left(y-2\right)}{15}=\frac{15x}{15}$  

 $\frac{\left(y-2\right)}{15}=x$  

 $B=5T+2h$  

let solve for T

 $B-2h=5T+2h-2h$  

This leaves us with  $B-2h=5T$  

 $\frac{\left(B-2h\right)}{5}=\frac{5T}{5}$  

 $\frac{\left(B-2h\right)}{5}=T$  

because of the parenthesis you have to get rid of the division of 2 by multiplying both sides by 2

 $2a=\frac{\left(4m+d\right)2}{2}$  

 $2a=4m+d$  

now that we got rid of the parenthesis we can now isolate d

 $2a-4m=4m-4m+d$  

 $2a-4m=d$  

In this formula you are dividing and multiplying and it doesn't matter which one you do first.

divide both sides by t or multiply both sides by 2

 $\frac{h}{t}=\frac{\left(\frac{1}{2}vt\right)}{t}\ or\ 2h=2\left(\frac{1}{2}vt\right)$  

then do the step you didn't do from above next

 $2\left(\frac{h}{t}\right)=2\left(\frac{1}{2}v\right)\ or\ \frac{2t}{t}=\frac{vt}{t}$