On putting x = -1 in x² + 2x + 1,we get
X² +2x+l=(-1)² +2(-1)+1
=1- 2 +1 = 2 -2 =0

On putting x = -1 in =:x + 2, we get
-x + 2 = -(-1) + 2=1 + 2=3

On putting x = -1 in 2x - 7, we get
2x - 7 = 2(-1) - 7 = - 2 - 7 = - 9

On putting p = -2 in -2p³ - 3p² +4p + 7 We get
-2p³ - 3p² + 4p + 7 = -2 (-2)³ - 3(-2)² + 4(-2)+ 7
- 2(-8) - 3(4) - 8 + 7 = 16 -12 - 8 + 7
16 + 7 - 12 - 8 = 23 - 20 = 3

On putting p = - 2 in -3p² + 4p + 7, we get
-3p² + 4p + 7 = -3(-2)² + 4(-2) + 7
= -3(4) -8 + 7= -12 -8 + 7
= -20 + 7= -13

On putting p = - 2 in 4p + 7, we get
4p + 7 = 4(-2) + 7 = - 8 + 7 = -1

On putting m = 2 in \(\frac { 5m }{ 2 } \) -4, we get
\(\frac { 5m }{ 2 } \) - 4= \(\frac { 5\times 2 }{ 2 } -4=\frac { 10 }{ 2 } \)- 4=5 - 4 = 1

On putting m = 2 in 3m² - 2m - 7, we get
3m² - 2m - 7 = 3(2)² - 2(2) - 7
= 3 X 4 - 4 - 7 = 12 -11 = 1

On putting m = 2 in 9 - 5m, we get
9 - 5m = 9 - 5 x 2 = 9 - 10 = -1

On putting m = 2 in 3m - 5, we get
3m - 5 = 3 X 2 - 5 = 6 - 5 = 1

On putting m = 2 in m - 2, we get
m- 2 = 2 - 2= 0

-11x² - 16xy +5y² + 10

5a³ - 6a² - 3a - 5

2x² - 6xy +4y² - x + y + 2

P³ + 9p² - 3p + 2

ab² + 2a²b + 4

6x - 2y+ 8z

a² - 7ab

3mn + 3

11xy - 6

11xy

Here, we have to subtract 4pq - 5q² - 5q^2 _-3p² from 5p² + 3q² + pq
 \(\quad 5p^2-3q^2-pq\\-3p^2-5q^2+4pq\\+\ \ \ \ \ \ \ +\ \ \ \ \ \ - \\\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\ \ \ 8p^2+8q^2-5pq\\\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\)

Here, We have to subtract 5a² -7ab +5b² from 3ab - 2a² - 2b²
 3ab  - 2a² - 2b²
-7ab + 5a² + 5b^{2
+        -           -
_________________}
10ab - 7a² -7b^{2
_________________}

Here, we have to subtract - x²  + 10x - 5 from 5x - 10
         5x - 10
-x² + 10x - 5
+     -      +
____________
  x² - 5x  - 5
____________

Here, We have to subtract -m² + 5mn from 4m² - 3mn + 8
\(4m^2-3mn+8\\-m^2+5mn\\-\ \ \ \ \ -\\\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\5m^2-8mn+8\\\_\_\_\_\_\_\_\_\_\_\_\_\_\_\)

We have to subtract, a(b - 5) from b(5 - a) or ab - Sa from 5b - ab
 \(\quad\quad \ 5b-ab\\-5a\quad+ab\\+\ \ \ \ \ \ \ \ +\\\_\_\_\_\_\_\_\_\_\_\_\_\\5a+5b-2ab\\\_\_\_\_\_\_\_\_\_\_\_\_\)

Here, we have to subtract (a - b) from (a + b).
\(\therefore\)  (a + b) - (a - b) = a-+ b - a + b = a - a + b + b
= a(1-1) + b(1+1)= a x 0 + b(2)
=0 + 2b = 2b

Here, we have to subtract 6xy from - 12xy
\(\therefore\)  (-12xy) - 6xy = -12xy - 6xy
= xy (-12- 6) = -18xy

Here, we have to subtract -5 y² from y²
\(\therefore\) y² - (-5y²) = y² + 5y² = y² (1+5) = 6y²

We have, (3y² + 5y - 4) - (8y - y² - 4)
= 3y²  + 5y - 4 - 8y + y² + 4
= 3y^{2 +} y² + 5y - 8y - 4 + 4  [rearranging terms]
= y² (3+1) + y (5-8)+ 0 -= 4y² - 3y

We have, 5x²y - 5x² + 3yx² - 3y² +x² - y² + 8xy² - 3y²
= 5x²y + 3yx² - 5x² + x² - x² - 3y² -y² - 3y²  + 8xy   [rearranging terms]
= x²y (5+3) + x² (-5 + 1) +y² (-3 -1 -3) + 8xy²
= 8x²y - 4x² - 7y² + 8xy

We have, 3a - 2b - ab - (a - b + ab) + 3ab + b - a
= 3a - 2b - ab - a + b - ab + 3ab + b - a
= 3a - a - a - 2b + b + b - ab - ab + 3ab  [rearranging terms]
= = a(3 -1 -1) + b(-2 + 1+ 1)+ ab(-1 -1 + 3)
= a + b x 0 + ab = a + ab

We have p- (p- q) - q- (q- p)
= p- p + q- q- q+ p
= p - p + p + q - q - q = p - q [rearranging terms)

We have -z² + 13z² - 5z + 7z³ -15z
= 7z³ - z² + 13z² - 5z - 15z   [rearranging terms]
= 7z³ + 12z² - 20z

We have 21b - 32 + 7b - 20b
= 21b + 7b - 20b - 32       [rearranging terms]
= 28b - 20b - 32 = 8b - 32  

If A= 3x²- 4x+ 1, B= 5x² + 3x- B,C= 4x² - 7x+ 3. Find the value of A + B - C
Then; \(\begin{matrix} A+B=3x^2-4x+1\\\quad \quad \quad\ \ \ \ 5x^2+3x-8\\ \quad \quad \quad \quad \quad\_\_\_\_\_\_\_\_\_\_\_\_ \\ \quad\ \ \ \ \ \ \ \ \ \ \ 8x^2-x-7\\ \quad\quad\quad\quad\quad\_\_\_\_\_\_\_\_\_\_\_\_ \end{matrix}\) \(\Rightarrow\) \(\begin{matrix} (A+B)-C=8x^2-x-7\\\quad \quad \quad\ \ \ \ 4x^2-7x+3\\\quad \quad -\ \ \ +\ \ \quad - \\ \quad \quad \quad \quad \quad\_\_\_\_\_\_\_\_\_\_\_\_ \\ \quad\ \ \ \ \ \ \ \ \ \ \ 4x^2+6x-10\\ \quad\quad\quad\quad\quad\_\_\_\_\_\_\_\_\_\_\_\_ \end{matrix}\)  [change the sign and then add]

Given, algebraic expressions are (x - y) and (x + y). So, subtraction of (x - y) from (x + y)
By horizontal method = (x + y) - (x - y) =x + Y - x + y = 2y
By column method,   
 x + y
-x \(\mp\) y
______
0 + 2y  or 2y
______

Given, algebraic expressions are (x - y) and (x + y).
By horizontal method, x - y + x + Y = x + x - y + Y = 2x
By column method,  
  x - y
+x + y
______
2x + 0 or 2x
_______

Coefficient x = 6xy²z² , Coefficient of xy = 6xyz² , coefficient of xyz = 6xyz

Two