-3/5 doesn't seem to be a solution to the question stem. Hence the only solution is X=1 making the answer C. Am i getting something wrong?

There are 3 steps to solving equations involving ABSOLUTE VALUE:
1. Apply the rule that says: If |x| = k, then x = k or x = -k
2. Solve the resulting equations
3. Plug solutions into original equation to check for extraneous roots


Given: |x + 3| = 4x
So, according to the above algorithm, EITHER x + 3 = 4x OR x + 3 = -4x
Let's examine each case...

Case i: x + 3 = 4x
Solve to get: x = 1
Plug x = 1 back into the original equation to get: |1 + 3| = 4(1)
Simplify both sides to get: |4| = 4. WORKS!
So, x = 1 is a valid solution


Case ii: x + 3 = -4x
Solve to get: x = -3/5
Plug x = -3/5 back into the original equation to get: |-3/5 + 3| = 4(-3/5)
Simplify both sides to get: |2 2/5| = -12/5. DOESN'T WORK
So, x = -3/5 is not a valid solution

So the ONLY valid solution is x = 1

We get:
QUANTITY A: 1
QUANTITY B: 1

Answer: C

Cheers,
Brent

You're correct.
I've edited the official answer to be C

Cheers,
Brent

|x + 3| = 4x

Quantity A	Quantity B
x	1

A Quantity A is greater.
B Quantity B is greater.
C The two quantities are equal.
D The relationship cannot be determined from the information given.

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There are 3 steps to solving equations involving ABSOLUTE VALUE:
1. Apply the rule that says: If |x| = k, then x = k or x = -k
2. Solve the resulting equations
3. Plug solutions into original equation to check for extraneous roots

So, here we get the following two equations: x + 3 = 4x and x + 3 = -4x
Let's solve each equation...

x + 3 = 4x
Subtract x from both sides of the equation: 3 = 3x
Solve: x = 1
To check for extraneous roots, plug x = 1 back into the original equation to get: |1 + 3| = 4(1). WORKS!
So, x = 1 is a solution.

x + 3 = -4x
Subtract x from both sides of the equation: 3 = -5x
Solve: x = -0.6
To check for extraneous roots, plug x = -0.6 back into the original equation to get: |(-0.6) + 3| = 4(-0.6).
Simplify: |2.4| = -2.4 FALSE
So, x = -0.6 is NOT a solution.

Since x = 1 is the only possible solution, the correct answer is C

|x + 3| = 4x

We will have to take two cases

Case 1: Whatever is inside the modulus is >= 0
=> x+3 >= 0 => x >= -3
=> |x+3| = x+3 (as |X| = X when X >= 0)
=> x+3 = 4x
=> 4x-x = 3
=> x = \(\frac{3}{3}\)= 1
1 >= -3
=> x = 1 is a solution

Case 2: Whatever is inside the modulus is < 0
=> x+3 < 0 => x < -3
=> |x+3| = -(x+3) (as |X| = -X when X < 0)
=> -(x+3) = 4x
=> x+3 = -4x
=>x + 4x = -3
=> 5x = -3
=> x = \(\frac{-3}{5}\)
But \(\frac{-3}{5}\) is not less than -3
=> So, \(\frac{-3}{5}\) is NOT a solution

=> x = 1

Clearly, Quantity A(x) = Quantity B (1)

So, Answer will be C
Hope it helps!

Watch the following video to learn how to Solve Inequality + Absolute value Problems

Can you please explain why are we checking for negative value of x?
Because if 4x is equal to mod of some equation then it is evident that x can only be positive. [because mod always gives positive value].
Therefore we need only check for positive x.

If |x| = k, then x = k or x = -k

See more on the module here and absolute value https://gre.myprepclub.com/forum/gre-ma ... 25236.html