How to Plug-in Difficult Options | Algebra

We all love to use the plug-in method in Quant questions. We have an equation given and if the options are the possible values of $x$ , we just plug in the values and find the value that satisfies the equation. But what if the options are complicated values? What if it seems that five times the calculation (in the worst-case) will be far more time consuming than actually solving the question? Then one is torn between using the favourite plug-in method and using algebra. Let’s take an example to review the methods we can use to solve the question and learn how to simplify the plug-in process by approximating the five available options.

Question: If ${|4x−4|=|2x+30|}$ , which of the following could be a value of ${x}$ ?

(A) ${\frac{-35}{3}}$

(B) ${\frac{−21}{2}}$

(D) ${\frac{11}{5}}$

(E) ${\frac{47}{5}}$

Solution: The question is an ideal candidate for “plug-in”. You have the absolute value equation in $x$ and the values of $x$ given in the options. The problem is that the values of $x$ given are fractional. Of course, if we do plan to solve the equation instead of plug-in here, we will solve it using our holistic approach rather than pure algebra. Let’s take a look at that and later we will discuss the trick to making the options easier for us.

Method 1: Solve for $x$ (method discussed in detail in our Inequalities module)

$|4x - 4| = |2x + 30|$

$4 \times |x - 1| = 2 \times |x + 15|$

$2 \times |x - 1| = |x + 15|$

This is how we rephrase the equation in our words – Twice the distance of $x$ from 1 should be equal to the distance of $x$ from -15.

——————(-15) —————————————————(0)——(1)——————

There are two ways to get the value of $x$ :

Case 1: $x$ could be between -15 and 1 such that the distance between them is split in the ratio 2 : 1.

Case 2: $x$ could be to the right of 1 such that the distance between $x$ and -15 is twice the distance between $x$ and 1.

Case 1:

The distance from -15 to 1 is of 16 units. Split it into 3 parts of $\frac{16}{3}$ each. So distance of $x$ from 1 should be $\frac{16}{3}$ and that will make the distance of $x$ from -15 twice of $\frac{16}{3}$ i.e. $\frac{32}{3}$ .
So $x$ should be at a point $\frac{16}{3}$ away from 1 toward the left.
$x = 1 - \frac{16}{3} = \frac{-13}{3}$

This is one of our options and hence the answer. Normally, we would just move on to the next question from this point. But had we not found $\frac{-13}{3}$ in options, we would have looked at case 2 too.

Case 2:

Distance between -15 and 1 is 16 units. $x$ should be further 16 units to the right of 1 so that distance between $x$ and 1 is 16 and the distance between $x$ and -15 is two times 16 i.e. 32. So $x$ should be 16 units to the right of 1 i.e. $x = 17$ .

If you would not have found $\frac{-13}{3}$ in the options, then you would have found 17.

Now let’s go on to see how we can make plug-in work for us in this case.

Method 2: Plug-in the options

$|4x - 4| = |2x + 30|$

$2 \times |x - 1| = |x + 15|$

Option (A) $\frac{-35}{3}$

It is difficult to solve for $x = \frac{-35}{3}$ to see if both sides match. Instead, let’s solve for the closest integer -12.

$2\times |-12 - 1| = |-12 + 15|$

On LHS, you get 26.

On RHS, you get 3.

The values are far away from each other and hence $x$ will not be $\frac{-35}{3}$ . As the value of $x$ approaches the point where the equation holds i.e. where the two sides are equal, the gap between the value of the two sides keeps reducing. With such a huge gap between the value of the two sides in this case, it is unlikely that a small adjustment of $\frac{-35}{3}$ from -12 will bring the two sides equal.

Option (B) $\frac{-21}{2}$

Solve for $x = -10$

$2 \times |-10 - 1| = |-10 + 15|$

On LHS, you get 22.

On RHS, you get 5.

The values are far away from each other and hence $x$ will not be $\frac{-21}{2}$

Option (C) $\frac{-13}{3}$

Solve for $x = -4$

$2 \times |-4 -1| = |-4 + 15|$

On LHS, you get 10.

On RHS, you get 11.

Here, there is a possibility that $x = \frac{-13}{3}$ . Plug in the actual value of $\frac{-13}{3}$ and you will see that LHS = RHS.
This is the answer.

Basically, we approximated the options and shortlisted the one that gave us very close values. We checked for that and found that it is the answer.

We can also solve this using pure algebra (taking positive and negative signs of absolute values) but in my opinion, the holistic method 1 is almost always better than that. Out of the two methods discussed above, you can pick the one you like better but note that method 2 has limited applications. Only if you are given the actual values of $x$ , you can use it. Method 1 is far more generic for absolute value questions.

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