What is the value of x?

When there’s a DS question on linear equations, be very mindful of the traps that might be laid out for the unsuspecting student. One of the most common traps is to make it seem like an equation can be solved when its actually not. Another is to give a set of dependent equations which don't yield unique solutions for the variable/s. Therefore, when you have DS questions on equations, your first line of attack should to be to figure out whether the equations are independent or not. If they are not, there's no point in wasting time on trying to solve them.

The data given in statement I is like that. It gives you an equation which cannot be solved beyond a certain stage.

From statement I alone,
\(\frac{(x+3) }{ 8}\) + \(\frac{(x-2) }{ 4}\) = \(\frac{(x+1) }{ 2}\) – \(\frac{(x+5) }{ 8}\).

Taking LCM on both sides and simplifying, we get,
\(\frac{(3x-1) }{ 8}\) = \(\frac{(3x – 1) }{ 8}\). We cannot solve this equation, this equation is like saying x=x.Knowing x=x is not sufficient to find A VALUE for x.

Statement I alone is insufficient. Answer options A and D can be eliminated. Possible answer options at this stage are B, C or E.

From statement II alone, \(x^2\) – 3x – 10 = x-14. Taking all the terms on to the LHS and simplifying, we get,

\(x^2\) – 4x +4 = 0 which can be written as \((x-2)^2\) = 0. This means that x=2. Statement II alone is sufficient to obtain a unique value for x.
Answer options C and E can be eliminated. The correct answer option is B.

Hope that helps!

Target question: What is the value of x?

Statement 1: \(\frac{x+3}{8} + \frac{x-2}{4} =\frac{ x+1}{2} – \frac{x+5}{8}\)
When studying for the GMAT's quantitative section, we soon learn that we should avoid performing unnecessary calculations when answering data sufficiency questions. So some students may look at the above equation and feel that, since they COULD solve the equation for x, then the statement must be sufficient. However this particular equation has infinitely many solutions. Let's see why:

Take: \(\frac{x+3}{8} + \frac{x-2}{4} =\frac{ x+1}{2} – \frac{x+5}{8}\)

Multiply both sides by 8 (the least common multiple of 2, 4 and 8) to get: \((x+3) + (2x-4) = (4x+4) - (x+5)\)
Simplify to get: \(3x-1=3x-1\)
At this point, we recognize that EVERY VALUE of x satisfies this equation.
In other words, x can equal ANY value.
As such, statement 1 is NOT SUFFICIENT

Statement 2: \(x^2 – 3x – 10 = x - 14\)
Add 14 to both sides of the equation to get: \(x^2 – 3x + 4 = x\)
Subtract x from both sides of the equation to get: \(x^2 – 4x + 4 = 0\)
Factor: \((x - 2)(x - 2)=0\)
So, it MUST be the case that x = 2
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: B

Cheers,
Brent