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Re: What is the value of integer x? [#permalink]
16 Jan 2013, 02:16

Expert's post

LM wrote:

What is the value of integer x?

(1) 2x^2+9<9x (2) |x+10|=2x+8

What is the value of integer x?

(1) 2x^2+9<9x --> factor qudratics: (x-\frac{3}{2})(x-3)<0 --> roots are \frac{3}{2} and 3 --> "<" sign indicates that the solution lies between the roots: 1.5<x<3 --> since there only integer in this range is 2 then x=2. Sufficient.

(2) |x+10|=2x+8 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: 2x+8\geq{0} --> x\geq{-4}, for this range x+10 is positive hence |x+10|=x+10 --> x+10=2x+8 --> x=2. Sufficient.

Re: M27-14 What is the value of integer x? [#permalink]
26 Apr 2013, 13:52

Expert's post

HumptyDumpty wrote:

My way: standard procedure: 1st: x+10=2x+8 x=2 OR 2nd:-(x+10)=2x+8 x=-6 the condition x=i is met: Insufficient.

This approach is NOT correct.

Always be sure to plug back the solutions for equations with modulus sign the values that you have found are x = 2 and x = -6 ONLY x=2 satisfies the given condition For X = -6

LHS = |x+10| = 4 RHS = 2x + 8 = -4 SO x = -6 is NO GOOD

This is happening because you have found the roots of (x+10)^2 = (2x+8)^2 _________________

Re: M27-14 What is the value of integer x? [#permalink]
26 Apr 2013, 13:57

Expert's post

This could ALSO be solved YOUR way but you have missed the definition of the modulus sign |x|= x WHEN x >or = 0 |x| = -x WHEN x<0

Now lets come back to the problem,

|x+10| = 2x+8

SO x+10 = 2x+8 when x+10> 0 So it gives x=2 Now lets check: is 2+10 > 0 YES. So this is GOOD Next -(x+10) = 2x+8 when x+10<0 it gives x=-6 Lets check: is -6+10 < 0 NO. So this is what you call an extraneous root. Does no good.
_________________

the sums in parentheses must have opposite signs, so:

\frac{3}{2}<x<3

consider the condition x=i:

x=2 Sufficient.

2) is not clear:

|x+10|=2x+8

My way: standard procedure: 1st: x+10=2x+8 x=2 OR 2nd:-(x+10)=2x+8 x=-6 the condition x=i is met: Insufficient.

The original explanation: (2) |x+10|=2x+8. The left hand side (LHS) is an absolute value, which is always non-negative, hence RHS must also be non-negative: 2x+8\geq0 giving us x\geq-4. Now, for this range x+10 is positive, hence|x+10|=x+10. So, |x+10|=2x+8 can be written as x+10=2x+8, solving for x gives x=2. Sufficient.

Noting that 2x+8\geq0 excludes the negative value and leaves off only one value. But what the heck is the mechanics behind this problem that makes the the good old way of solving inequalities insufficiently precise here?

Merging similar topics.

The way you call "standard procedure" is not complete.

When expanding |x+10|=2x+8 you should consider x<-10 range and x>=-10 range:

x<-10 --> -(x+10)=2x+8 --> x=-6. Discard this solution since it's not in the range x<-10. x\geq{-10} --> x+10=2x+8 --> x=2.

Re: What is the value of integer x? [#permalink]
10 Mar 2014, 04:47

Bunuel wrote:

LM wrote:

What is the value of integer x?

(1) 2x^2+9<9x (2) |x+10|=2x+8

What is the value of integer x?

(1) 2x^2+9<9x --> factor qudratics: (x-\frac{3}{2})(x-3)<0 --> roots are \frac{3}{2} and 3 --> "<" sign indicates that the solution lies between the roots: 1.5<x<3 --> since there only integer in this range is 2 then x=2. Sufficient.

(2) |x+10|=2x+8 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: 2x+8\geq{0} --> x\geq{-4}, for this range x+10 is positive hence |x+10|=x+10 --> x+10=2x+8 --> x=2. Sufficient.

Re: What is the value of integer x? [#permalink]
10 Mar 2014, 05:15

Expert's post

sanjoo wrote:

Bunuel wrote:

LM wrote:

What is the value of integer x?

(1) 2x^2+9<9x (2) |x+10|=2x+8

What is the value of integer x?

(1) 2x^2+9<9x --> factor qudratics: (x-\frac{3}{2})(x-3)<0 --> roots are \frac{3}{2} and 3 --> "<" sign indicates that the solution lies between the roots: 1.5<x<3 --> since there only integer in this range is 2 then x=2. Sufficient.

(2) |x+10|=2x+8 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: 2x+8\geq{0} --> x\geq{-4}, for this range x+10 is positive hence |x+10|=x+10 --> x+10=2x+8 --> x=2. Sufficient.