Why is nothing being done to right side numerator?

Question

$$\frac{3(t-7)}{2} = t-6$$

From what I understand is that you have to multiply both sides by $2$, so that on the left side $2$ cancel out and you are left with $3(t-7)$.

but why does right side turn into $2t-12$?

So now you have: \begin{gather*} 3(t-7) = 2t-12\\ 3t-21=2t-12\\ t-21=-12\\ t=9\\ \end{gather*}

I think it's right.

Answer 1

$\frac{3(t-7)}{2}=t-6$

$2\times \frac{3(t-7)}{2}=2\times (t-6)$ multiply both sides by 2

$\frac{2[3(t-7)]}{2}=2[(t-6)]$ we see here that the 2 in the numberator and the 2 in the denominator in the LHS can cancel out so we get,

$3t-21=2(t)-2(6)$ distribute the 2 on the RHS and the 3 on the LHS

$3t-21=2t-12$

$3t-2t-12=-12$ subtract 2t from both sides

$t=-12+21$ add 21 to both sides

$t=9$

Details: $\frac{2(3(t-6))}{2}$

$\frac{2(3t-18)}{2}$

$\frac{6t-36}{2}$

but $\frac{6t}{2}=3t$ and $\frac{36}{2}=18$

Answer 2

The solution you give is correct.

When you multiply $t-6$ by $2$, you get $2(t-6)$, and it is because of the distributive law that that becomes $(2t) - (2\cdot6)$, which is $2t-12$.