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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