# Mathematics - INEQUALITY Short Tricks

**INEQUALITY**

Before going through the
questions, we need to understand these terms in inequality.

**Variables-**

A symbol which can be
assume various numerical values is called variable.

Example: 5x + 7y = 40 here x, y are variables.

**Constants-**

A symbol which has fixed
value is called constant.

Example: 7x + 3y + 20z = 44 here is 7, 3, 20, 44 are constants values.

**Coefficient-**

Any factor of a term is called the coefficient of the remaining factor
of the term.

Example: 7x+3y= 19 here 7 is a coefficient of x and 3 is of y.

**Now have a look how to find the value of variables.**

**Case: 1**

**7x + 3y = 15……. Eq 1**

**10x + 5y = 10……. Eq 2**

Solution:

At first we need the
same coefficient of either x or y, so we multiply the eq- 1 by 5 & eq- 2 by
3.

7x + 3y = 15 (Multiply by 5)

10x + 5y = 10 (Multiply by 3)

After multiplying we get
these two eq. Now solve them by subtracting eq- 1 by eq- 2.

35x + 15y = 75

30x + 15y = 30

We get

5x = 45

x = 9

Put this value of x in
any eq- and find out value of y.

7 x 9 + 3y = 15

y = 17 Approx

So, x = 9 and y = 17

**Case: 2**

**i- 6x^2 + x - 1 = 0**

**ii- 8y^2 + 10y + 3 =0**

Solution:

In equation- 1: +6 is
coefficient of x2

+1 is coefficient of x

-1 is constant term

1: we multiply (+6) x
(-1) = -6

2: we break 6 in two
parts such that subtract (because constant term sign is (-)) of these part
should be 1.

So, -6 = 3 - 2 = 1 and
product of both factors is 6

3: Change the sign of
both the factors, So +3 = -3 and -2 = +2 and

Divide by coefficient of
x^2, So we get -3/6 = -1/2 and

+2/6 = +1/3

We get x = -1/2 and +1/3

Similarly y = -3/4 and
-1/2

Now other
things we need to learn is comparison of these values and find the right
answer. In
quadratic Equations we have the following options to choose from:-

i. X >Y
(x is greater than y or y is less then x)

ii. Y
>X (x is less than y or y is greater than x)

iii. X
>= Y (x is greater than or equal to y)

iv. Y>=
X (x is less than or equal to y)

v. X = Y (x
is equal to y) or relationship cannot be established.

(i) Now, Suppose on
solving the quadratic equation we get X = -1, 2 And Y = -2, -3

On putting these values
on number line we will see that X lies on the right of Y. Hence, We can say
that X>Y or Y<X.

(ii) Now, Suppose if
after solving the equation we get X = -1 , 2 and Y= 4, 5

On putting these values
on number line we see that Y lies to the right of X without touching X at any
point. Hence, We can say that Y>X or X<Y.

(iii) If X= 3, 4 and Y=
1, 3. Putting this on number line we get X to the right of Y but touching Y at
one single point i.e., 3. Hence, X>=Y or Y<=X.

(iv) If X=3, 4 and Y=4,
6. Putting these values on number line we will see that Y lies to the right of
X but touching X at 4. Hence, Y>=X or X<=Y.

(v) Now, Suppose if y=7
and X= 4, 8 then X<y also X>y. Hence, in this case no relationship can be
established.

So according
to our example in case – 2

X= -1/2 and 1/3

Y= -3/4 and -1/2

Here we can see that X=Y
and X>Y, so our right option is X>=Y i.e. opt (iii)

**Note – If you still feel trouble leave comment.**

**Keep practice ....Thank You.**

**Solve Q- 1 to 5 and your options are...**

(i). If x>y

(ii).
If x>=y

(iii).
If x<y

(iv).
If x<=y

(v).
x=y or relationship cannot be established

1) (a) X^2 + X-20= 0 and

(b) Y^2-Y-30= 0.

2) (a) 225X^2-4 = 0 and

(b)
(225y)^(1/2) +2 = 0

3) (a) 2x^2 + 11x + 14 = 0 and

(b) 4y^2 + 12y +9 =0

4) (a) x^4- 227= 398 and

(b) y^2 + 321=346

5) (a) x^2-7x+12=0 and

(b) y^2-12y+32=0

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