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# INEQUATIONS(Inequalities) Part 1

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Narenn

Example 3 :- $$\frac{(2x+4)}{(x-1)}$$ ≥ 5

$$\frac{(2x+4)}{(x-1)}$$ ≥ 5

$$\frac{(2x+4)}{(x-1)}$$ - 5 ≥ 0

$$\frac{((2x+4)-5(x-1))}{(x-1)}$$ ≥ 0

$$\frac{(-3x+9)}{(x-1)}$$ ≥ 0

$$\frac{(3x-9)}{(x-1)}$$ ≤ 0 ----------------[ Multiplying both sides by -1]

$$\frac{(3(x-3))}{(x-1)}$$ ≤ 0

$$\frac{(x-3)}{(x-1)}$$ ≤ 0 -----------------[ dividing both sides on Inequation by 3]

x = 3, 1

Sign of original Inequation obtained in Step 4 is negative so the solution set will be union of the regions containing negative sign.

Therefore Solution Set is 1 < x ≤ 3

NOTE :-
When x=1, the Inequation becomes $$\frac{6}{0}$$ ≥ 5 and we know that division by zero is not defined, hence we need to exclude x=1 from solution set.

Great work!

Shouldn't the highlighted portion be 1 ≤ x ≤ 3
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Example 3 :- $$\frac{(2x+4)}{(x-1)}$$ ≥ 5

$$\frac{(2x+4)}{(x-1)}$$ ≥ 5

Great work!

Shouldn't the highlighted portion be 1 ≤ x ≤ 3

No, because (x-1) in denominator will yield zero at X=1 and answer undefined.
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PiyushK
Game
Example 3 :- $$\frac{(2x+4)}{(x-1)}$$ ≥ 5

$$\frac{(2x+4)}{(x-1)}$$ ≥ 5

Great work!

Shouldn't the highlighted portion be 1 ≤ x ≤ 3

No, because (x-1) in denominator will yield zero at X=1 and answer undefined.

Example 4 :- \frac{(x+3)}{(x-2)} ≤ 2

Why based on the same concept, 2 has not been excluded in example 4? In this example also, 2 in the denominator will result the expression to undefined
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PiyushK
Game
Example 3 :- $$\frac{(2x+4)}{(x-1)}$$ ≥ 5

$$\frac{(2x+4)}{(x-1)}$$ ≥ 5

Great work!

Shouldn't the highlighted portion be 1 ≤ x ≤ 3

No, because (x-1) in denominator will yield zero at X=1 and answer undefined.

How come when you multiply the equation by (x-1), you don't eliminate it from the denominator?
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Hi Sir,
In Example 4, the answer reads as: x ≥ 7 or x ≤ 2

Shouldn't the answer be x ≥ 7 or x < 2
The reason being that if x=2, then the denominator becomes 0, and we can't have that.
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Narenn
Hi Sir,
In Example 4, the answer reads as: x ≥ 7 or x ≤ 2

Shouldn't the answer be x ≥ 7 or x < 2
The reason being that if x=2, then the denominator becomes 0, and we can't have that.

I believe you're correct here.

As a rule of thumb, shouldn't the test taker always test the inequality solution to ensure 0 doesn't result in the denominator?
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ak1802
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Hi Sir,
In Example 4, the answer reads as: x ≥ 7 or x ≤ 2

Shouldn't the answer be x ≥ 7 or x < 2
The reason being that if x=2, then the denominator becomes 0, and we can't have that.

I believe you're correct here.

As a rule of thumb, shouldn't the test taker always test the inequality solution to ensure 0 doesn't result in the denominator?

Hi ak1802 and kanav06,

yes both of you are correct, the values of x will not include x=2..

say for an Inequality you are getting the range of x between 0 and 7, 0<x<7, BUT in the denominator of that fraction is x-2 ----------- abc/(x-2) >0.....
the range will become 0<x<7, where $$x\neq{2}$$...
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ak1802
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Narenn
Hi Sir,
In Example 4, the answer reads as: x ≥ 7 or x ≤ 2

Shouldn't the answer be x ≥ 7 or x < 2
The reason being that if x=2, then the denominator becomes 0, and we can't have that.

I believe you're correct here.

As a rule of thumb, shouldn't the test taker always test the inequality solution to ensure 0 doesn't result in the denominator?

Hi ak1802 and kanav06,

yes both of you are correct, the values of x will not include x=2..

say for an Inequality you are getting the range of x between 0 and 7, 0<x<7, BUT in the denominator of that fraction is x-2 ----------- abc/(x-2) >0.....
the range will become 0<x<7, where $$x\neq{2}$$...

Hi chetan2u,

Say we have a $$\frac{(x+3)(x-2)}{x(x-1)}$$ <0

How do we represent this number line. Esp how do we represent x on number line. Clearly x cannot be =0 and -1.

then on number line how can we get the solution for this inequality.

Here is what i did
$$-3 -1 0 2$$
<-----|------------|-----------|-----------------|---------------->
+VE -VE +VE -VE +VE

so the solution can be -3<x<-1 and 0<x<2

is this correct approach .

I want to understand how do we represent 'x" if we have something like x(x-a)(x-b) . I am not sure how do we get to this .

Probus
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Bunuel Narenn Can someone help with @probus's question?
I have the same doubt.
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INEQUATIONS (INEQUALITIES): Prologue

Example 4 :- $$\frac{(x+3)}{(x-2)}$$ ≤ 2

1. $$\frac{(x+3)}{(x-2)}$$ ≤ 2

2. $$\frac{(x+3)}{(x-2)}$$ - 2 ≤ 0

3. $$\frac{(x+3-2x+4)}{(x-2)}$$ ≤ 0

4. $$\frac{(-x+7)}{(x-2)}$$ ≤ 0

5. $$\frac{(x-7)}{(x-2)}$$ ≥ 0 -------------------[Multiplying both sides by -1]

x-7 ≥ 0 ---------> x ≥ 7

x-2 ≥ 0 ---------> x ≥ 2

x = 7 and 2 are the critical points.

The Real line is divided into three regions. Since the Inequation (obtained in step 4) possesses greater than sign which means that LHS of the Inequation is positive. So, the solution set of the given Inequation is the union of the regions containing positive sign.

Hence x ≥ 7 or x ≤ 2 -----------> (-∞, 2] U [7, ∞)

---------- End of First Part ------------

Quick questions about Example 4 / inequalities in general, and also looking for clarification. I added Step #'s beside each line in the quoted example, just for clarity when writing this up.

1) Between steps 1. and 2., if we had left 2 on the RHS, we couldn't multiply the entire inequality by (x-2) because it could be negative, correct? But we can manipulate by bringing the 2 over, then multiply 2 by (x-2) to create 1 fraction with a common denominator, and not affect the overall inequality? (i hope that makes sense - I might be overthinking this).

2) Between steps 4. and 5., we're allowed tomultiply the numerator by -1, and therefore flip the inequality, because we know the inequality in Step 4 is < 0 (so we know the sign), right?
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