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Re: |x+3|=4x [#permalink]
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-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?

Originally posted by smgbendi on 07 Jun 2020, 15:08.
Last edited by smgbendi on 08 Jun 2020, 12:33, edited 1 time in total.
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|x+3|=4x [#permalink]
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Asmakan wrote:
\(|x+3|=4x\)


Quantity A
Quantity B
x
1



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
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Re: |x+3|=4x [#permalink]
smgbendi wrote:
-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?


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

Cheers,
Brent
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|x + 3| = 4x [#permalink]
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|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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Re: |x + 3| = 4x [#permalink]
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|x + 3| = 4x [#permalink]
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NickOP wrote:
|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.


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
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Re: |x + 3| = 4x [#permalink]
1
|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

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Re: |x + 3| = 4x [#permalink]
GreenlightTestPrep wrote:
arpitjain wrote:
|x + 3| = 4x

Quantity A
Quantity B
x
1



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, if |x + 3| = 4x, we get: x + 3 = 4x and x + 3 = -4x

Let's solve each equation.
Take: x + 3 = 4x
Subtract x from both sides: 3 = 3x
Solve: x = 1

Plug x = 1 into original equation to get: |1 + 3| = 4(1)
Simplify: 4 = 4. Works!!
So, x = 1 is a VALID solution
----------------------------------

Take: x + 3 = -4x
Subtract x from both sides: 3 = -5x
Solve: x = -3/5

Plug x = -3/5 into original equation to get: |-3/5 + 3| = 4(-3/5)
Simplify: 12/15 = -12/15. DOESN'T WORK
So, x = -3/5 is NOT a valid solution
----------------------------------------

Since x = 1 is the ONLY valid solution, we get:
QUANTITY A: 1
QUANTITY B: 1

Answer: C

Cheers,
Brent


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.
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Re: |x + 3| = 4x [#permalink]
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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
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Re: |x + 3| = 4x [#permalink]
1
By solving the equation we get x=1, -3/5.
By putting the x's values on the equation we find that x=1 makes both sides equal which is 4.
but x=-3/5 makes left hand 12/5 and right hand side -12/5 which are not equal. So -3/5 can not be the value of x. x=1 only.
Correct answer C.
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Re: |x + 3| = 4x [#permalink]
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