@xiggi - I would say that the manipulations are not trivial. Three algebraic steps at the least, and all the while, it is not certain that you are getting anywhere! That was my point: we have not seen enough new SAT problems to know that the approach is going to be worth the time. If you had told me that this was a current-SAT problem, I definitely would pursue your approach. In fact, with a little tweaking (so that the answer would not be negative) and this would be a fine grid-in question.
@Xiggi I’m not sure I understand your first solution - perhaps you can give a few more details? But your second solution is good. Here are the details:
(1) We subtract the first equation from the second to get 4y - 3x = -2 - 4x + 2y
(2) Solve for x to get x = -2 - 2y = -2(1 + y) = -2(y + 1)
(3) Cross divide to get (y + 1)/x = -1/2, choice (B).
I suppose that you could argue that there are only 2 steps here, but I personally see 3 steps: (i) Performing an operation (subtraction), (ii) Solving for a variable (in this case x), and (iii) Recognizing that the expression in (ii) can be manipulated into the required form.
I suppose that you can argue that (ii) and (iii) can be considered one step, but even so, I think you would agree that it would still be more difficult than anything we’ve seen on the current SAT.
Also, I don’t know if I would recommend this method here. What if we were asked to find a different final expression. I don’t think this method would work for an arbitrary expression.
I’ll post a few more solutions for this problem soon, but anyone else please feel free to post your own in the meantime.
By the way, the method of “performing a simple operation” (as Xiggi suggested and I gave details for in my last post) will still be an effective strategy on the new SAT (according to the CB’s released material), but it will no longer work for almost every problem.
^^
OK, here’s what is happening on my paper and it might explain the different counts on steps. I realize I did collapse steps as I wrote down the two equations (to help the subtracting) as follows:
3x =5 - 5y
2x =7 - 3y
I admit that I did evaluate the given equations and was looking for a way to isolate a 1x on the left side and “hopefully” find a (y+1) on the other side. What I did get is x = -2 +2y
My next lines could have been x = -2(y+1) or -1/2x = -1/2 (-2+2y)
My “one” step solution stems from making those manipulations in my head (as I see them) but I understand that one considers those as additional steps. Anyhow, why I considered the -1/2 as the only answer is because from looking at the line above, I knew that the (y+1)/x has to have a negative solution. I did not consider the -11 as a viable solution and I eliminated the two positive answers. On this note, I think that something can be learned from the “arcane” methods used by good test writers. Although I cannot demonstrate this with great effectiveness, I think that one could see a “hint” in the proposed answers that points to 2 and -1/2 being the “plausible answers” as they are somehow opposites and could yield to both the correct answer for the astute and the erroneous one for the careless student. I find this behavior common on the SAT – well the older versions that is.
For all intents and purposes, were I to have to explain to a student how to approach this problem, it would be a better idea to follow the methods described above, and not try to explain my “intuitive” or “visual” methods that rely on the type of equation balancing my Eurodad taught me. His words were “switch sides and change the sign.” You need to add the accent for full effect. 
Not sure any of this is really helpful, but I tried to explain how my mind works! One thing, however, that I think is essential in the problem is to keep the focus on the “solution” that includes the full term (y+1)/x as opposed to solve for both x and y.
Are there any books or websites to practice for the new sat?
Ok - I promised some solutions to this problem. Here is the first one:
2x + y = 7 – 2y
5y – x = 5 – 4x
If (x,y) is a solution to the above system of equations, what is the value of (y+1)/x ?
(A) -11
(B) -1/2
© 2
(D) 20
Solution using the elimination method: We begin by making sure that the two equations are “lined up” properly. We do this by adding 2y to each side of the first equation, and adding 4x to each side of the second equation.
2x + 3y = 7
3x + 5y = 5
We will now multiply each side of the first equation by 5, and each side of the second equation by -3.
5(2x + 3y) = (7)(5)
-3(3x + 5y) = (5)(-3)
Do not forget to distribute correctly on the left. Add the two equations.
10x + 15y = 35
- 9x – 15y = -15
x = 20
Using the first equation in the solution to find y, we have
2⋅20 + 3y = 7
40 + 3y = 7
3y = 7 – 40 = -33
y = (-33)/3 = -11
So (y+1)/x = (-11+1)/20 = -10/20 = -1/2, choice (B).
**Remarks:/b We chose to use 5 and -3 because multiplying by these numbers makes the y column “match up” so that when we add the two equations in the next step the y term vanishes. We could have also used -5 and 3.
(2) If we wanted to find y first instead of x we would multiply the two equations by 3 and -2 (or -3 and 2). In general, if you are only looking for one variable, try to eliminate the one you are not looking for. In this case we need to find both so it doesn’t matter which we find first.
(3) We chose to multiply by a negative number so that we could add the equations instead of subtracting them. We could have also multiplied the first equation by 5, the second by 3, and subtracted the two equations, but a computational error is more likely to occur this way.
Solution using Gauss-Jordan reduction: As in the first solution (elimination method), we first make sure the two equations are “lined up” properly.
2x + 3y = 7
3x + 5y = 5
We’ll use our TI-84 Calculator now. Begin by pushing the MATRIX button. Scroll over to EDIT and then select A. We will be inputting a 2×3 matrix, so press 2 ENTER 3 ENTER. We then begin entering the numbers 2, 3, and 7 for the first row, and 3, 5, and 5 for the second row. To do this we can simply type 2 ENTER 3 ENTER 7 ENTER 3 ENTER 5 ENTER 5 ENTER.
Note: What we have just done was create the augmented matrix for the system of equations. This is simply an array of numbers which contains the coefficients of the variables together with the right hand sides of the equations.
Now push the QUIT button (2ND MODE) to get a blank screen. Press MATRIX again. This time scroll over to MATH and select rref( (or press B). Then press MATRIX again and select A and press ENTER.
Note: What we have just done is put the matrix into reduced row echelon form. In this form we can read off the solution to the original system of equations.
Warning: Be careful to use the rref( button (2 r’s), and not the ref( button (which has only one r).
The display will show the following.
[ [1 0 20]
[0 1-11]]
The first line is interpreted as x = 20 and the second line as y = -11.
So we have
(y + 1)/x = (-11 + 1)/20 = -10/20 = -1/2, choice (B).
^^ Inasmuch as the solutions work, I still think that it is NOT necessary to solve for x and y separately, and this because of the problem statement that asks for “(y + 1)/x”
I’d say that the students who follow HS math principles are heading for a time sink for this particular problem. The time wasted going through the additional steps solving this easy problem might cost a student 2 more unanswered questions.
@Xiggi If this were for the current SAT I would agree with you completely. Since this is for the new SAT, my feeling is that it is not a certainty that a single operation will produce the answer right away. (In this problem it does, but in another similar problem it might not.)
I’m not saying not to use the method you described for this particular problem - it certainly works here.
But I think for the new test advanced students should practice the standard methods as well - elimination, Guass-Jordan reduction, substitution, and various graphing calculator methods.
Let me add that we won’t know for sure if the method of “trying a simple operation” will always work for these problems until we see a few actual new SATs. The evidence suggests it will not. For example, the College Board released a sample question involving a similar system of equations that asks for the value of xy. There is no simple manipulation that will find xy without finding x and y individually first.
To be clear, I am not sure the definitions of a simple operation versus a series of short steps really matter. What I call a simple operation might be viewed by others as two or three. My point is all about first looking at the proposed answers (since it eliminates two) and most importantly looking at the format of the answer, which was not about solving for neither x nor y.
It might have been a coincidence that the short step that includes the subtraction yielded an equation that was easily transformed into the format required, but I considered it a … brilliant move by the test writer. Brilliant because it opened the trap to someone blindly following the lengthy path as well as gave sufficient ammunition to the astute test taker to pick the right (and only plausible) answer.
As far as what comes in the future, I am happy to agree with the notion that the SAT will probably turn into more of a formulaic and “dumber” math test as it will follow the Common Core. The intuitive and more mentally agile test taker will find a lot fewer low hanging fruits in the next versions.
But then, “we” have been rewarding mediocrity in education for several decades in a departure from the ROW.
PS To be clear, I do not consider the graphical method in 46 to be dumber. Not at all!