# Why is nothing being done to right side numerator?

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:

3(t-7) = 2t-12

3t-21=2t-12

t-21=-12

t=9

I think it's right.

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

$2\times \frac{3(t-7)}{2}=2\times (t-6)$ 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 $\frac{36}{2}=18$

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I see how they cancel out on LHS and not on RHS because on RHS denominator is 1 not 2!? I think that's partial explanation that would make sense to me... –  Liger86 Sep 28 '11 at 3:16
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 $2t-12$.