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If |x+1|=2|x-1|, x=?

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If |x+1|=2|x-1|, x=?

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[Math Revolution GMAT math practice question]

If |x+1|=2|x-1|, x=?

1) x<1
2) x>0

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Max@Math Revolution wrote:
[Math Revolution GMAT math practice question]

If |x+1|=2|x-1|, x=?

1) x<1
2) x>0
Let´s use the geometric interpretation of the absolute value, so that NO calculations will be needed!
$$\left. \matrix{
\left| {x - 1} \right|\,\, = {\rm{dist}}\left( {x,1} \right) \hfill \cr
\left| {x + 1} \right| = \left| {x - \left( { - 1} \right)} \right|\,\,{\rm{ = }}\,\,{\rm{dist}}\left( {x, - 1} \right)\,\,\, \hfill \cr} \right\}\,\,\,\,\,\,\mathop \Rightarrow \limits_{{\rm{given}}}^{{\rm{equation}}} \,\,\,\,\,dist\left( {x, - 1} \right)\,\,\, = \,\,\,2 \cdot dist\left( {x,1} \right)$$



FOCUS : x

Important: without retrictions imposed in the statements, we already know there are EXACTLY two values of x that satisfy the equation given.

(1) Only in the first figure we have x less than 1, hence SUFFICIENT.

(2) There is no need to obtain x in the first figure (although it is obviously 1 - [1-(-1)]/3 = 1/3) to realize it is positive.
(Zero is the average of -1 and 1, and 0 is to the left of the red dot, "evident" even without precision in the figure created.)

Hence the two red dots are viable: (2) is INSUFFICIENT.

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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Max@Math Revolution wrote:
[Math Revolution GMAT math practice question]

If |x+1|=2|x-1|, x=?

1) x<1
2) x>0
(I gave some time to other people present an alternate solution... it didn´t happen.)

Let´s offer an alternate approach:
$$? = x$$
$$\left. \matrix{
{\left| {x + 1} \right|^{\,2}} = {\left( {x + 1} \right)^{\,2}} \hfill \cr
{\left| {x - 1} \right|^{\,2}} = {\left( {x - 1} \right)^{\,2}}\,\,\,\, \hfill \cr} \right\}\,\,\,\,\,\,\mathop \Rightarrow \limits_{{\rm{equation}}\,\,{\rm{given}}}^{{\rm{squaring}}\,\,{\rm{the}}} \,\,\,\,\,\,{\left( {x + 1} \right)^2} = 4{\left( {x - 1} \right)^2}\,\,\,\, \Rightarrow \,\,\,\, \ldots \,\,\,\, \Rightarrow \,\,\,\,3{x^2} - 10x + 3 = 0$$
$$3{x^2} - 10x + 3 = 0\,\,\,\,\,\mathop \Rightarrow \limits_{{\text{product = 1}}}^{{\text{sum}}\,{\text{ = }}\frac{{{\text{10}}}}{3}} \,\,\,\,\,\boxed{\,\,\,x = \frac{1}{3}\,\,\,\,{\text{or}}\,\,\,x = 3\,\,\,\,}\,\,\,\left( * \right)\,\,\,\,\,$$
Important: whenever an equation is squared - or put to any positive EVEN power - solutions are never lost, but new ones may be (undesirably) created.
That´s why all roots obtained (after the procedure) must be checked ("tested") in the original equation. Both values are solutions to the original equation.
$$\left( 1 \right)\,\,x < 1\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,x = \frac{1}{3}\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{SUFF}}.$$
$$\left( 2 \right)\,\,x > 0\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,x = \frac{1}{3}\,\,\,\,{\text{or}}\,\,\,x = 3\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{INSUFF}}.\,$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

_________________
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br



Last edited by fskilnik@GMATH on Mon Oct 15, 2018 2:44 pm; edited 1 time in total

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Max@Math Revolution wrote:
[Math Revolution GMAT math practice question]

If |x+1|=2|x-1|, x=?

1) x<1
2) x>0
Case 1: Signs unchanged
x+1 = 2(x-1)
x+1 = 2x-2
3 = x

Case 2: Signs changed on ONE SIDE
-x-1 = 2(x-1)
-x-1 = 2x-2
1 = 3x
1/3 = x

Resulting solutions:
x=3 or x=1/3

Statement 1:
Since x < 1, only x=1/3 is a viable solution.
SUFFICIENT.

Statement 2:
Since x>0, both x=1/3 and x=3 are viable solutions.
Since x can be different values, INSUFFICIENT.

The correct answer is A.

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=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

The original condition |x+1|=2|x-1| is equivalent to x=1/3 or x=3 as shown below:

|x+1|=2|x-1|
=> |x+1|^2=(2|x-1|)^2
=> (x+1)^2=4(x-1)^2
=> x^2+2x+1=4(x^2-2x+1)
=> x^2+2x+1=4x^2-8x+4
=> 3x^2-10x+3 = 0
=> (3x-1)(x-3) = 0
=> 3x-1=0 or x-3 = 0
=> x=1/3 or x=3

Thus, condition 1) is sufficient since it gives a unique solution.

Condition 2) is not sufficient, since it allows both possible solutions.

Therefore, A is the answer.
Answer: A

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