When the unknown shows up on both sides of the equals sign, you cannot just undo operations in order. You first have to gather all the variable terms on one side. The SAT builds two kinds of questions on this: ordinary solving, and questions where the variables cancel completely and the answer is about how many solutions exist.
Take 5x plus 8 equals 2x minus 7. Subtract 2x from both sides to get 3x plus 8 equals negative 7, then subtract 8 to get 3x equals negative 15, so x equals negative 5. The order matters less than the discipline: move all the x terms one way and all the plain numbers the other way, then divide once at the end.
Choose the side that keeps the coefficient positive. In the equation above, subtracting 5x instead would give 8 equals negative 3x minus 7, which ends at the same place but carries a negative coefficient through two more lines. Moving the smaller variable term is a small habit that removes a lot of sign errors.
Parentheses come first. In 4 times the quantity x minus 3 equals 2x plus 10, distribute to get 4x minus 12 equals 2x plus 10, then 2x equals 22 and x equals 11. The distribution error to watch for is a negative in front of parentheses. In 9 minus 3 times the quantity x plus 2, the second term becomes minus 3x minus 6, not minus 3x plus 6, because the 3 being distributed is negative.
Now the interesting case. Simplify 6x plus 4 equals 6x minus 1. Subtracting 6x from both sides leaves 4 equals negative 1, which is false no matter what x is, so the equation has no solution. If instead you reach something like 4 equals 4, the statement is true for every x, so every real number is a solution. These two outcomes are the whole point of the questions that include an unknown constant.
That is how constant questions work. In 6x plus 4 equals kx minus 1, ask when the variables cancel: that needs k equal to 6, and then the leftover statement 4 equals negative 1 is false, so k equals 6 gives no solution. For infinitely many solutions you would need both the coefficients and the constants to match, which the negative 1 prevents here.