Is there an error in the algebra of this GMAT problem explanation? A problem I found on this site is stated as follows:

Determine the value of $t$.
  
  
*
  
*$2t + 6s = 8$
  
*$t/2 - 2 = - 3s/4$

One is supposed to determine whether or not the problem can be solved by the given conditions (1 and 2).
Here is the official explanation of the answer:
Correct answer is: Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

Correct answer: Statements (1) and (2) TOGETHER are NOT sufficient. In
  statement 1, $2t + 6s = 8$ is one equation is two unknowns, hence we
  cannot determine the value of t. The statement is insufficient. In
  statement 2, $t/2 - 2 = -3s/4$ can be transformed to $t - 4 = -6s/2$. But
  this is an equation with two unknowns hence we cannot determine the
  value of $t$. The statement is insufficient. Combining the two
  statements, we have  2, $t/2 - 2 = -3s/4$ which can be transformed to $t
 - 4 = -6s/2$ then to $2t - 8 = -6s$ but this is equal to the equation $2t + 6s = 8$, hence we have an equation with two unknowns. Thus, we cannot determine the value of $t$; the statements (1) and (2) TOGETHER are NOT
  sufficient.

My problem is with this statement:

In statement 2, $t/2 - 2 = -3s/4$ can be transformed to $t - 4 = -6s/2$.

If both sides are multiplied by $4$, shouldn't it be $2t - 8 = -3s$?
In the end, my answer was 

BOTH statements (1) and (2) TOGETHER are sufficient to answer the
  question asked, but NEITHER statement ALONE is sufficient to answer
  the question ask

Which is correct?
 A: You are obviously correct : the equations, treated together, admit the solution $t =4 , s= 0$.
The mistake is in the part indicated : indeed, the "can be transformed" part is incorrect. However, note that if $2.$ were changed to $2. \frac t2 - 2 = -\frac{3s}2$ then in fact said transformation can be completed. It seems to be an error of computation or in the original question, in that case.
I just went through the entire test on your site, and yes, it is a bad error which needs correction.
EDIT : For completeness, I will add Dave Renfro's comment. Given one equation in two unknowns, it is in fact possible that both unknowns can be found from that equation alone. For example, $s^2+t^2 = 0$ is uniquely possible only with $s=t=0$ (over reals, or rationals/integers if you like). Even under the restriction of this being linear, $3(t-2s)+6s = 6$ is satisfied by a unique $t$ and any $s$, so one variable is still getting fixed.
The necessary and sufficient condition for two equations of the form $ax+by + c = 0$ and $dx+ey+f = 0$ to be simultaneously satisfied by a unique pair of $x,y$ is that $ae \neq bd$. In your case, you have $2t+6s - 8 = 0$ and $t/2 + 3s/4 - 2 = 0$. We have $a=2,b=6,d=\frac 12, e = \frac 34$ ,so $ae = 1.5$ and $bd = 3$. These are not equal, giving a unique satisfying pair.
Suppose $ae = bd$ , then $\frac{a}{d} = \frac{b}{e}$ (assuming $d,e \neq 0$ , else the second equation will be degenerate). If $c=f=0$ or if $f \neq 0$ and $\frac{c}{f}$ equals those fractions above, then the two equations have infinitely many simultaneous solutions. Else, they don't have any simultaneous solution.
