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Re: If ax + b = 0, is x > 0 ? (1) a + b > 0 (2) a - b > 0 [#permalink]
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Bunuel wrote:
uzzy12 wrote:
If ax + b = 0, is x > 0

(1) a + b > 0
(2) a - b > 0


Given: \(b=-ax\). Question: is \(x>0\)

(1) \(a+b>0\) --> \(a-ax>0\) --> \(a(1-x)>0\) --> either \(a>0\) and \(1-x>0\), so \(x<1\) OR \(a<0\) and \(1-x<0\), so \(x>1\). Not sufficient.

(2) \(a-b>0\) --> \(a+ax>0\) --> \(a(1+x)>0\) --> either \(a>0\) and \(1+x>0\), so \(x>-1\) OR \(a<0\) and \(1+x<0\), so \(x<-1\). Not sufficient.

(1)+(2) Sum (1) and (2) (we can do this as the signs of these inequalities are in the same direction) --> \((a+b)+(a-b)>0\) --> \(a>0\), so we have first range from (1): \(x<1\) and first case from (2): \(x>-1\) --> \(-1<x<1\), so \(x\) may or may not be negative. Not sufficient.

Answer: E.



Bunuel Please help me in understanding this by (1)+(2) we came up to \(a>0\), now from question we have \(-b=ax\) which means product of ax is negative and by that logic would not x be <0 as we have got \(a>0\).... Please help me in understanding flaw in my logic
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Re: If ax + b = 0, is x > 0 ? (1) a + b > 0 (2) a - b > 0 [#permalink]
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sonnco wrote:
ax + b = 0, is x > 0?
1. a + b > 0
2. a - b > 0
How is the OA answer right?! Your input is greatly appreciated.


\(ax+b=0\)
\(ax=-b\)
\(x=-\frac{b}{a}\)

\(Is \hspace{3} x > 0\)
\(Is \hspace{3} -\frac{b}{a} > 0\)
\(Is \hspace{3} \frac{b}{a} < 0\)

OR

Does a and b have opposite signs?

1. \(a + b > 0\)
\(a>-b\)

It doesn't tell us anything about sign.
a=5
b=2
a>-b as 5>-2
Same sign. No.

a=5
b=-2
a>-b as 5>2
Opposite signs. Yes.

Not Sufficient.

2. \(a - b > 0\)
\(a>b\)

It doesn't tell us anything about sign.
a=5
b=2
a>b as 5>2
Same sign. No.

a=5
b=-2
a>b as 5>-2
Opposite signs. Yes.

Combining both;

a > |b|

Using same sample set:
a=5, b=2, No.
a=5, b=-2, Yes.
Not Sufficient.

Ans: "E"
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Re: If ax + b = 0, is x > 0 ? (1) a + b > 0 (2) a - b > 0 [#permalink]
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Bunuel wrote:
uzzy12 wrote:
If ax + b = 0, is x > 0

(1) a + b > 0
(2) a - b > 0


Given: \(b=-ax\). Question: is \(x>0\)

(1) \(a+b>0\) --> \(a-ax>0\) --> \(a(1-x)>0\) --> either \(a>0\) and \(1-x>0\), so \(x<1\) OR \(a<0\) and \(1-x<0\), so \(x>1\). Not sufficient.

(2) \(a-b>0\) --> \(a+ax>0\) --> \(a(1+x)>0\) --> either \(a>0\) and \(1+x>0\), so \(x>-1\) OR \(a<0\) and \(1+x<0\), so \(x<-1\). Not sufficient.

(1)+(2) Sum (1) and (2) (we can do this as the signs of these inequalities are in the same direction) --> \((a+b)+(a-b)>0\) --> \(a>0\), so we have first range from (1): \(x<1\) and first case from (2): \(x>-1\) --> \(-1<x<1\), so \(x\) may or may not be negative. Not sufficient.

Answer: E.

you've made things easier bunuel. +1
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Re: If ax + b = 0, is x > 0 ? (1) a + b > 0 (2) a - b > 0 [#permalink]
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awesome explanations both bunuel and mainhoon --these two approaches would probably serve two different types of students really well. nice work!

there's also a potential blend of the two-- mainhoon's rephrase (are the signs of a and b different) and the stacking/adding of inequalities when evaluating choice c for that same rephrased question (if your head gets turned around when too many numbers are involved)
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Re: If ax + b = 0, is x > 0 ? (1) a + b > 0 (2) a - b > 0 [#permalink]
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sonnco wrote:
ax + b = 0, is x > 0?

1. a + b > 0
2. a - b > 0



You can look at this geometrically - because the equation is that of a line, the question is set up so that you can look at it as a coordinate geometry question. The equation y = ax + b is the equation of a line with slope a and y-intercept b. When we plug in y=0, we are finding the x-intercept of the line. So the question is just asking if the x-intercept of the line y=ax + b is positive. In a diagram, that would be true if our line crosses the x-axis to the right of the origin (0,0).

Now, using both statements, if a > -b and a > b, then a > |b|. So a is clearly positive, and our line has a positive slope and is climbing as you move to the right. We can now just draw two different scenarios on the coordinate plane, one where b is positive and one where b is negative. If you choose, say, b = 1 and a = 5, then if you quickly sketch your line you'll see that the x-intercept of y = ax + b is negative, and if b=-1 and a = 5 then if you quickly sketch your line you'll see that the x-intercept of y=ax+b is positive, so not sufficient, and the answer is E.
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Re: If ax + b = 0, is x > 0 ? (1) a + b > 0 (2) a - b > 0 [#permalink]
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given \(ax=- b\) then x = \(\frac{-b}{a}\) question becomes is \(\frac{-b}{a}>0\)

From statement 1

\(a+b > 0\) so the values for \((a,b)>> (+,+) , (-,+), (+,-)\)

\(a-b>0\) so the values for \((a,b) >> (+,+), (+,-)\)

Combination still 2 cases remain. Hence E

Is this method safe?
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Re: If ax + b = 0, is x > 0 ? (1) a + b > 0 (2) a - b > 0 [#permalink]
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fozzzy wrote:
given \(ax=- b\) then x = \(\frac{-b}{a}\) question becomes is \(\frac{-b}{a}>0\)

From statement 1

\(a+b > 0\) so the values for \((a,b)>> (+,+) , (-,+), (+,-)\)

\(a-b>0\) so the values for (a,b) >> (+,+), (+,-)

Combination still 2 cases remain. Hence E

Is this method safe?


Actually, all the possible cases are : (+,+) = 7,5 ; (+,-) = 7,-5 ; (-,-) = -5,-7.

Otherwise, IMO it is all good.
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Re: If ax + b = 0, is x > 0 ? (1) a + b > 0 (2) a - b > 0 [#permalink]
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uzzy12 wrote:
If ax + b = 0, is x > 0

(1) a + b > 0
(2) a - b > 0


Guessing numbers helped me.

x = -b/a
Is -b/a > 0?

(1) a > -b
a = 5, b =3 => N
a = 5, b=-3 => Y
NS
(2) a > b
Same numbers as in (1)
NS

(1) + (2)
a > |b|
Same numbers used in (1) and in (2)
NS
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Re: If ax + b = 0, is x > 0 ? (1) a + b > 0 (2) a - b > 0 [#permalink]
we have a+b>0
a-b>0
if we add both the equations. a>0
if we subtract both the equations, b>0
then we know both have same signs....
hence can determine, sign of x=-b/a is >0 or not
What is wrong in my explanation?
some please help...
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sudd1 wrote:
we have a+b>0
a-b>0
if we add both the equations. a>0
if we subtract both the equations, b>0
then we know both have same signs....
hence can determine, sign of x=-b/a is >0 or not
What is wrong in my explanation?
some please help...
Bunuel


For two inequalities, you can only apply subtraction when their signs are in the opposite directions:
If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from).
Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

So, we cannot subtract a+b>0 from a-b>0.

For more check: inequalities-tips-and-hints-175001.html
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Re: If ax + b = 0, is x > 0 ? (1) a + b > 0 (2) a - b > 0 [#permalink]
Would this approach be wrong?

Choose two numbers: a=3, b=2.
a+b>0 and a-b>0 are individually and together fulfilled.

ax+b= 3*x+2=0 => x=-2/3 < 0 => we have shown that both I and II are insufficient.
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Re: If ax + b = 0, is x > 0 ? (1) a + b > 0 (2) a - b > 0 [#permalink]
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meltedcheese wrote:
Would this approach be wrong?

Choose two numbers: a=3, b=2.
a+b>0 and a-b>0 are individually and together fulfilled.

ax+b= 3*x+2=0 => x=-2/3 < 0 => we have shown that both I and II are insufficient.


To get insufficiency you should get both a NO and an YES answers to the question. You have a NO answer, so to get that the statements are insufficient, you should choose numbers which give an YES answer to the question.
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Re: If ax + b = 0, is x > 0 ? (1) a + b > 0 (2) a - b > 0 [#permalink]
I am getting 2 different equations here.. Please help where I am going wrong..

ax + b = 0

So X=-b/a


1) a+b>0

So, a>-b--- for a>0

1>-b/a
1>x
or -a>-b--- for a<0
1<-x
x<-1


2) a-b>0

a>b---- for a>0
1<b/a
x<-1

-a>b--- for a<0
1<-b/a
1<x
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Re: If ax + b = 0, is x > 0 ? (1) a + b > 0 (2) a - b > 0 [#permalink]
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Dear sheolokesh

The mistake you made was at the part highlighted in red:

sheolokesh wrote:

1) a+b>0

So, a>-b--- for a>0

1>-b/a
1>x

or -a>-b--- for a<0
1<-x
x<-1




Correct processing of the case a < 0 would be as follow:

From Statement 1,

a + b > 0
--> a > - b . . . (1)

Case: a < 0

Dividing both sides of an inequality with a negative number changes the sign of inequality.

So, dividing both sides of Inequality 1 with a, we get:

1 < \(\frac{-b}{a}\)

Substituting -b/a = x, we get:

1 < x
That is, x > 1

So, from Statement 1, we see that

If a > 0, then x < 1
And, if a < 0, then x > 1

So, we cannot say for sure if x is positive or not.

Similarly, in your analysis of Statement 2, you got confused between the impact of a being positive or negative on the sign of inequality:

sheolokesh wrote:

2) a-b>0

a>b---- for a>0
1<b/a Correct expression: 1 > b/a
x<-1 Correct expression: 1 > -x
Correct expression: -1 < x, that is, x > -1


-a>b--- for a<0 Correct expression: a > b
1<-b/a Correct expression: 1 < b/a (multiplying both sides by negative number changes sign)
1<x Correct expression: 1 < -x
Correct expression: -1 > x, that is, x < -1 (multiplying both sides by negative number changes sign)



It is okay to make mistakes as long as we learn from them. And, the important takeaway from our discussion of this mistake is:

Don't skip steps when multiplying or dividing terms on both sides of an inequality. Because, this step is particularly prone to errors.

Hope our discussion was helpful! :)

Best Regards

Japinder
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Re: If ax + b = 0, is x > 0 ? (1) a + b > 0 (2) a - b > 0 [#permalink]
Bunuel wrote:
uzzy12 wrote:
If ax + b = 0, is x > 0

(1) a + b > 0
(2) a - b > 0


Given: \(b=-ax\). Question: is \(x>0\)

(1) \(a+b>0\) --> \(a-ax>0\) --> \(a(1-x)>0\) --> either \(a>0\) and \(1-x>0\), so \(x<1\) OR \(a<0\) and \(1-x<0\), so \(x>1\). Not sufficient.

(2) \(a-b>0\) --> \(a+ax>0\) --> \(a(1+x)>0\) --> either \(a>0\) and \(1+x>0\), so \(x>-1\) OR \(a<0\) and \(1+x<0\), so \(x<-1\). Not sufficient.

(1)+(2) Sum (1) and (2) (we can do this as the signs of these inequalities are in the same direction) --> \((a+b)+(a-b)>0\) --> \(a>0\), so we have first range from (1): \(x<1\) and first case from (2): \(x>-1\) --> \(-1<x<1\), so \(x\) may or may not be negative. Not sufficient.

Answer: E.


Bunuel

If I find from the question stem itself that x is negative i.e. x=-b/a, in this case, can we say we dont require any additional statements to prove it and hence, E?
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Re: If ax + b = 0, is x > 0 ? (1) a + b > 0 (2) a - b > 0 [#permalink]
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sinhap07 wrote:
Bunuel wrote:
uzzy12 wrote:
If ax + b = 0, is x > 0

(1) a + b > 0
(2) a - b > 0


Given: \(b=-ax\). Question: is \(x>0\)

(1) \(a+b>0\) --> \(a-ax>0\) --> \(a(1-x)>0\) --> either \(a>0\) and \(1-x>0\), so \(x<1\) OR \(a<0\) and \(1-x<0\), so \(x>1\). Not sufficient.

(2) \(a-b>0\) --> \(a+ax>0\) --> \(a(1+x)>0\) --> either \(a>0\) and \(1+x>0\), so \(x>-1\) OR \(a<0\) and \(1+x<0\), so \(x<-1\). Not sufficient.

(1)+(2) Sum (1) and (2) (we can do this as the signs of these inequalities are in the same direction) --> \((a+b)+(a-b)>0\) --> \(a>0\), so we have first range from (1): \(x<1\) and first case from (2): \(x>-1\) --> \(-1<x<1\), so \(x\) may or may not be negative. Not sufficient.

Answer: E.


Bunuel

If I find from the question stem itself that x is negative i.e. x=-b/a, in this case, can we say we dont require any additional statements to prove it and hence, E?


No.

1. You cannot write x = -b/a from ax + b = 0, because a can be 0 and division by 0 is not allowed.
2. Even if we had x = -b/a, minus sign does not indicate that x is negative. If b/a is negative itself then x = -b/a = -(negative) = positive.

Hope it's clear.
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Re: If ax + b = 0, is x > 0 ? (1) a + b > 0 (2) a - b > 0 [#permalink]
Bunuel wrote:
uzzy12 wrote:
If ax + b = 0, is x > 0

(1) a + b > 0
(2) a - b > 0


Given: \(b=-ax\). Question: is \(x>0\)

(1) \(a+b>0\) --> \(a-ax>0\) --> \(a(1-x)>0\) -->either \(a>0\) and \(1-x>0\), so \(x<1\) OR \(a<0\) and \(1-x<0\), so \(x>1\). Not sufficient.

(2) \(a-b>0\) --> \(a+ax>0\) --> \(a(1+x)>0\) --> either \(a>0\) and \(1+x>0\), so \(x>-1\) OR \(a<0\) and \(1+x<0\), so \(x<-1\). Not sufficient.

(1)+(2) Sum (1) and (2) (we can do this as the signs of these inequalities are in the same direction) --> \((a+b)+(a-b)>0\) --> \(a>0\), so we have first range from (1): \(x<1\) and first case from (2): \(x>-1\) --> \(-1<x<1\), so \(x\) may or may not be negative. Not sufficient.

Answer: E.


Bunuel I have a doubt in equality:
a>0 and 1−x>0, so x<1 OR a<0 and 1−x<0, so x>1. not able to get how did a>0 turned to a<0 and x<1 to 1-x<0
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