Negative Numbers

Using the principle, for any two positive numbers x,y, we write,

-x + y = y-x

⇒ 5-10 = -5

Note: The sign of bigger number will come in the result.

Using the principle, for any two positive numbers x,y, we write,

-x + y = y-x

⇒ 5-10 = -5

Note: The sign of bigger number will come in the result.

Using the principle, for any two positive numbers x,y, we write,

-x + y = y-x

-10 + 5 can be written as,

⇒ 5-10

⇒ 5-10 = -5

Note: The sign of bigger number will come in the result.

Using the principle, For any two positive numbers x,y, we write,

-x-y = -(x + y)

⇒ -(5 + 10) = -(15)

Note: If both numbers are negative then result also will be negative.

Using the principle, For any two positive numbers x,y, we write,

-x-y = -(x + y)

⇒ -(5 + 5) = -10

Note: If both numbers are negative then result also will be negative.

Using the principle, for any two positive numbers x,y, we write,

-x + y = y-x

⇒ 5-5 = 0

Note: A number and it’s negative addition result in null and also called additive inverse.

⇒

⇒

⇒1

Note: The sign of bigger number will come in the result.

⇒

⇒

⇒ -2

Note: If both fractions are negative then result also will be negative.

⇒

⇒ = 0

Note: if a fraction and its negative are added then their result is 0.

Let take as positive numbers as: 1,2,3,4

Negative numbers: - 1,-2,-3,-4 and 0

For positive number:

(x + 1) = 1 + 1, 2 + 1 ,3 + 1 ,4 + 1

= 2,3,4,5

(x-1) = 1-1, 2-1 ,3-1 ,4-1

= 0, 1, 2, 3

(1-x) = 1-1, 1-2, 1-3, 1-4

= 0, -1, -2, -3

(i) (1 + x) + (1-x) = 2

Let’s check for x = 1

L.H.S = (1 + 1) + (1-1)

= 2 + 0

= 2

= R.H.S

Hence, it works for x = 2

Let’s check for x = 3

L.H.S = (1 + 3) + (1-3)

= 4 – 2

= 2

= R.H.S

Hence, it works for x = 3

Similarly, we can check for other values too. It will follow.

x – (x – 1) = 1

Let’s check for x = -1

L.H.S = -1-(-1-1)

= -1-(-2)

= 1

= R.H.S

Hence, it works for x = -1

Let’s check for x = -2

L.H.S = -2 - (-2-1)

= -2 – (-3)

= 1

= R.H.S

Hence, it works for x = -2

Let’s check for x = -3

L.H.S = -3 – (-3 - 1)

= -3 – (-4)

= 1

= R.H.S

Hence, it works for x = -3

Similarly we can check for other values too. It will follow.

1 – x = – (x – 1)

Let’s check for x = 0

L.H.S = 1 – 0

= 1

R.H.S = -(0 - 1)

= 1

= L.H.S

Hence, it works for x = 0

Let’s check for x = 1

L.H.S = 1 – 1

= 0

R.H.S = - (1 - 1)

= 0

= L.H.S

Hence, it works for x = 1

Let’s check for x = 2

L.H.S = 1 – 2

= -1

R.H.S = - (2 - 1)

= -1

= L.H.S

Hence, it works for x = 2

Similarly we can check for other values too. It will follow.

Let’s take x as : 1,2,3,4

And respectively y as : -1,-2,-3,-4

For x = 1 and y = -1:

(x + y) = (1 - 1) = 0

(x - y) = (1 – (-1)) = 2

For x = 2 and y = -2:

(x + y) = (2 - 2) = 0

(x - y) = (2 – (-2)) = 4

For x = 3 and y = -3:

(x + y) = (3 - 3) = 0

(x - y) = (3 – (-3)) = 6

For x = 4 and y = -4

(x + y) = (4 - 4) = 0

(x - y) = (4 – (-4)) = 8

(i) (x + y) – x = y

Let’s check for x = 1 and y = -1

L.H.S = (1 + (-1)) – 1

= (0) – 1

= -1

R.H.S = y

= -1

= L.H.S

Hence, it holds for x = 1 and y = -1

Let’s check for x = 2 and y = -2

L.H.S = (2 + (-2)) – 2

= (0) – 2

= -2

R.H.S = -2

= L.H.S

Hence, it works for x = 2 and y = -2

Similarly we can check for other values too. It will follow.

(x + y) – y = x

Let’s check for x = 3 and y = -3

L.H.S = (3 + (-3)) – (-3)

= (0) + 3

= 3

R.H.S = 3

= L.H.S

Hence, it works for x = 3 and y = -3

Let’s check for x = 4 and y = -4

L.H.S = (4 + (-4)) – (-4)

= (0) + 4

= 4

R.H.S = 4

= L.H.S

Hence, it works for x = 4 and y = -4

Similarly, we can check for other values too. It will follow.

(x – y) + y = x

Let’s check for x = 1 and y = -1

L.H.S = (1 – (-1)) + (-1)

= (2) – 1

= 1

R.H.S = 1

= L.H.S

Hence, it works for x = 1 and y = -1

Let’s check for x = 2 and y = -2

L.H.S = (2 – (-2)) + (-2)

= (4) -2

= 2

R.H.S = 2

= L.H.S

Hence, it works for x = 2 and y = -2

Similarly we can check for other values too. It will follow.

Let’s take as x = 1,2,3,4,5 (positive numbers)

And as x = -1,-2,-3,-4,-5 (negative numbers)

(i) –x + (x + 1) = 1

We will calculate LHS in every case and compare it with RHS taking x = 1, = -1 + (1 + 1) = -1 + 2

= 1

Taking x = 2, = -2 + (2 + 1) = -2 + 3

= 1

taking x = 3,

= -3 + (3 + 1)

= -3 + 4 = 1

taking x = 4,

= -4 + (4 + 1)

= -4 + 5 = 1

taking x = 5,

= -5 + (5 + 1)

= -5 + 6 = 1

now taking x as negative numbers Taking x = -1, = -(-1) + (-1 + 1)

= 1 – 1 + 1 = 1

Taking x = -2, = -(-2) + (-2 + 1)

= 2 + 1 -2 = 1

Taking x = -3, = -(-3) + (-3 + 1)

= 3 -3 + 1 = 1

Taking x = -4, = -(-4) + (-4 + 1)

= 4 -4 + 1 = 1

Taking x = -5, = -(-5) + (-5 + 1)

= 5 – 5 + 1 = 1

As in each case LHS = 1 so, LHS = RHS

Hence, above equation is an identity.

–x + (x + 1) + (x + 2) – (x + 3) = 0

We will calculate LHS in every case and copare it with LHS

Taking x = 1,

= -1 + (1 + 1) + (1 + 2)-(1 + 3)

= -1 + 2 + 3-4

= 0

Taking x = 2,

= -2 + (2 + 1) + (2 + 2)-(2 + 3)

= -2 + 3 + 4-5

= 0

Taking x = 3,

= -3 + (3 + 1) + (3 + 2)-(3 + 3)

= -3 + 4 + 5-6

= 0

Taking x = 4,

= -4 + (4 + 1) + (4 + 2)-(4 + 3)

= -4 + 5 + 6-7

= 0

Taking x = 5,

= -5 + (5 + 1) + (5 + 2)-(5 + 3)

= -5 + 6 + 7-8

= 0

Taking x = -1,

= -(-1) + (-1 + 1) + (-1 + 2)-(-1 + 3)

= 1 + 0 + 1-2

= 0

Taking x = -2,

= -(-2) + (-2 + 1) + (-2 + 2)-(-2 + 3)

= 2-1 + 0-1

= 0

Taking x = -3,

= -(-3) + (-3 + 1) + (-3 + 2)-(-3 + 3)

= 3-2-1 + 0

= 0

Taking x = -4,

= -(-4) + (-4 + 1) + (-4 + 2)-(-4 + 3)

= 4-3-2 + 1

= 0

Taking x = -5,

= -(-5) + (-5 + 1) + (-5 + 2)-(-5 + 3)

= 5-4-3 + 2

= 0

As in each case in LHS = 0 so, LHS = RHS

Hence, above equation is an identity.

–x – (x + 1) + (x + 2) + (x + 3) = 4

We will calulate LHS in every case and compare it with RHS

Taking x = 1,

= -1-(1 + 1) + (1 + 2) + (1 + 3)

= -1-2 + 3 + 4

= 4

Taking x = 2,

= -2-(2 + 1) + (2 + 2) + (2 + 3)

= -2-3 + 4 + 5

= 4

Taking x = 3,

= -3-(3 + 1) + (3 + 2) + (3 + 3)

= -3-4 + 5 + 6

= 4

Taking x = 4,

= -4-(4 + 1) + (4 + 2) + (4 + 3)

= -4-5 + 6 + 7

= 4

Taking x = 5,

= -5-(5 + 1) + (5 + 2) + (5 + 3)

= -5-6 + 7 + 8

= 4

Taking x = -1,

= 1-(-1 + 1) + (-1 + 2) + (-1 + 3)

= 1 + 0 + 1 + 2

= 4

Taking x = -2,

= 2-(-2 + 1) + (-2 + 2) + (-2 + 3)

= 2 + 1 + 0 + 1

= 4

Taking x = -3,

= 3-(-3 + 1) + (-3 + 2) + (-3 + 3)

= 3 + 2-1 + 0

= 4

Taking x = -4,

= 4-(-4 + 1) + (-4 + 2) + (-4 + 3)

= 4 + 3-2-1

= 4

Taking x = -5,

= 5-(-5 + 1) + (-5 + 2) + (-5 + 3)

= 5 + 4-3-2

= 4

As in each case in LHS = 4 so, LHS = RHS

Hence, above equation is an identity.

Let’s take x = 0,1,-1 y = 1,0,-1 and z = -1,1,0 respectively.

CASE 1: When x = 0, y = 1 and z = -1

Then, x + (y + z),

= 0 + (1-1)

= 0

CASE 2: When x = 1, y = 0 and z = 1

Then, x + (y + z)

= 1 + (0 + 1)

= 2

CASE 3: When x = -1, y = -1 and z = 0,

Then, x + (y + z)

= -1 + (-1 + 0)

= -2

Calculating (x + y) + z

CASE 1: When x = 0, y = 1 and z = -1

= (0 + 1) + (-1)

= 0

CASE 2:When x = 1, y = 0 and z = 1

= (1 + 0) + 1

= 2

CASE 3:When x = -1, y = -1 and z = 0

= (-1-1) + 0

= -2

Since, in every case x + (y + z) = (x + y) + z so,this holds for all these numbers.

let, take x = 0,1,-1 y = 1,0,-1 and z = -1,1,0 respectively.

CASE 1:When x = 0, y = 1, and z = -1,

(x + y)z = (0 + 1)-1

= 1×-1

= -1

CASE 2:When x = 1, y = 0, and z = 1,

(x + y)z = (1 + 0)1

= 1×1

= 1

CASE 3:When x = -1, y = -1, and z = 0,

(x + y)z = (-1-1)0

= -2×0

= 0

Calculating xz + yz

CASE 1:when x = 0, y = 1, and z = -1,

xz + yz = 0×-1 + 1×-1

= 0 - 1

= -1

CASE 2:When x = 1, y = 0 and z = 1,

xz + yz = 1×1 + 0×1

= 1 + 0

= 1

CASE 3: When x = -1, y = -1 and z = 0

Xz + yz = -1×0 + (-1×0)

= 0 + 0

= 0

As in all case (x + y)z = xz + yz hence, it will holds for all numbers.

1. y = x², x = –5, x =2. 5

Putting x = -5,

y = (-5)

² = -5×-5 = 25

Putting x = 5,

y = (5)² = 5×5

= 25

y = x² + 3x + 2, x = –2

Putting y = -2,

Y = (-2)² + 3×(-2) + 2

= 4-6 + 2

= 0

y = x² + 5x + 4, x = –2, x = –3

When x = -2,

y = (-2)² + 5×-2 + 4

= 4-10 + 4

= -2

When x = -3,

y = (-3)² + 5×-3 + 4

= 9-15 + 4

= -2

y = x³ + 1, x = –1

Putting x = -1,

y = (-1)³ + 1

= 0

y = x³ + x² + x + 1, y = –1

putting y = -1,

y = (-1)³ + (-1)²-1 + 1

y = -1 + 1 -1 + 1

= 0

s = 12t – 2t²

(i) when t = 6, s = 12×6-2(6)²

= 72-36×2

= 0

Thus, it is neither left nor right of point P.

(ii) s = 2t(6-t)

At t = 6s,

s = 2×6(6-6) = 0

Hence, it is at point P.

(iii) s = 12t- 2t² it can have written as 2t(6-t)

After 6 seconds the term (6-t) will be negative

At t = 7 s = 2×7(6-7)

s = 14 ×-1

s = -14

At t = 8 s = 2×8(6-8)

s = 16× -2

s = -32

It will be left of point P after t = 6 seconds

We have to find how many pairs of integers are there, satisfying the equation, x² + y² = 25

When x = 0, 0 + y² = 25 so, y = + 5,-5

Same will be for when y = 0, x² + 0 = 25 so, x = 5,-5

By hit and trial we can also find 9 + 16 or 16 + 9 = 25

So, x = 3,-3,4,-4 and respective y = 4,-4,3,3

So pairs are (0,5),(0,-5),(5,0),(-5,0),(3,4),(-3,4),(4,3),(4,-3),(-4,-3),(-3,-4).

y = and x =

when x = y = so, y = -

when x = - y = so, y = -2

when x = - y = so, y = -

When x = -2 : y =

So, y = + (-1) y = -1- = -

When x = - : y = so, y =

So, y = + 2 =

When x = 10, y = –5

Then, z = - so, z = -2 +

z =

when x = –10, y = 5

then, z = - so, z = -2 +

z =

when x = –10, y = –5

then, z = - so, z = 2-

z =

when x = 5, y = –10

then, z = - so, z = 2-

z =

when x = –5, y = 10

then, z = - so, z = -2-

z =