Isolating a Variable

[Jason76]

\(\displaystyle 5z + j =r(z - 9)\)

z = ?

Maybe this is the way:

\(\displaystyle \frac{5z + j}{(z - 9)} = r\)

Is this a good start?

 

[MarkFL]

I would first distribute on the right to get:

\(\displaystyle 5z+j=rz-9r\)

Now, arrange the equation with the terms having z as a factor on the left, and everything else on the right:

\(\displaystyle rz-5z=j+9r\)

Now, factor z out on the left:

\(\displaystyle (r-5)z=j+9r\)

Now, divide through by \(\displaystyle r-5\) to get:

\(\displaystyle z=\dfrac{j+9r}{r-5}\)

 

[Jason76]

I would first distribute on the right to get:

\(\displaystyle 5z+j=rz-9r\)

Now, arrange the equation with the terms having z as a factor on the left, and everything else on the right:

\(\displaystyle rz-5z=j+9r\)

Now, factor z out on the left:

\(\displaystyle (r-5)z=j+9r\)

Now, divide through by \(\displaystyle r-5\) to get:

\(\displaystyle z=\dfrac{j+9r}{r-5}\)

Thanks for the help

 

[Deleted member 4993]

Don't forget the constraint → \(\displaystyle r \ne 5\)