We have:
(14x + 10y - 12xy - 13) + (18 - 7x - 10y + 8xy) + 4xy
= 14x + 10y - 12xy - 13 + 18 - 7x - 10y + 8xy + 4xy
= (14x - 7x) + (10y - 10y) + (-12xy + 8xy + 4xy) + (-13 + 18)
= (14 - 7)x + (10 - 10)y + (-12 + 8 + 4)xy + (5)
= (7)x + (0)y + (-12 + 12)xy + 5
= 7x + 0y + (0)xy + 5
= 7x + 5

a + b - 3, b - a + 3, a - b + 3
We have 
(a + b - 3) + (b - a + 3) + (a - b + 3)
=a+b-3+b-a+3+a-b+3
= (a - a + a) + (b + b - b) + (-3 + 3 + 3)
= (1 - 1 + l)a + (1 + 1 - 1)b + (-3 + 6)
= (2 - 1)a + (2 - l)b + (-3 + 6)
= (1)a + (1)b + (3)
=a+b+3

The required expression is obtained by subtracting (4b² - a²) from (b² - 4a²).
∴ The required expression
= (b² - 4a²) - (4b² - a²)
= b² - 4a² - 4b² + a²
= (b² _ 4b²) + (-4a² + a²)
= (1 - 4)b² + (-4 + 1)a²
= (-3)b² + (-3)a²
= -3b² - 3a²
Thus, the required expression is (-3b² - 3a²).

The required expression is obtained by
subtracting (4a² - b + 2) from (3a² - 3b + 6).
∴ We have (3a² - 3b + 6) - (4a² - b + 2)
= 3a² - 3b + 6 - 4a² + b - 2
= (3a² - 4a²) + (-3b + b) + (6 - 2)
= (3 - 4)a² + (-3 + l)b + (4)
= -a² - 2b + 4
Thus, the expression (-a² - 2b + 4) can be subtracted from 3a² - 3b + 6 to get 4a² - b + 2.

Sum of 5x² - 6x + 4 and -x²r - 2x + 3
= (5x² - 6x + 4) + (-4x² - 2x + 3)
= 5x² - 6x + 4 - 4x² - 2x + 3
= (5x² - 4x²) + (-6x - 2x) + (4 + 3)
= (5 - 4)x² + (-6 - 2)x + (7)
= (I)x² + (-8x) + 7
=x²-8x+7
Now subtract x² - 8x + 7 from 0, we get 
∴  0 - [x² - 8x + 7] = 0 - x² + 8x - 7
= -x² + 8x - 7

A-B = (3c² - 4b - 1) - (6c²+ 3b - 8)
= 3a² - 4b - 1 - 6c² - 3b + 8
= (3a² - 6c²) + (-4b - 3b) + (-1 + 8)
= (3 - 6)a² + (-4 - 3)b + (-1 + 8)
= (-3)a² + (-7)b + (7)
= -3a² - 7b + 7
∴ A - B + C = (-3a² - 7b + 7) + (4a² - 9b + 3)
= -3a² - 7b + 7 + 4a² - 9b + 3
= (-3a² + 4a²) + (-7b - 9b) + (7 + 3)
= (-3 + 4)a² + (-7 - 9)b + (10)
= (1)a² + (-16)b + 10
= a² - 16b + 10

(a)Total amount deposited in account = (5x² + 3x-2) + (x²-x + 5)
=5x² + 3x-2+ x²-x + 5
=5x² + x²+ 3x-x-2+ 5
=6x²+2x+3
Balance amount
= (6x²+2x+3)-(2x²-x-3)
= 6x²+2x+3-2x²-x-3
= 6x²-2x²+ 2x + x + 3 + 3
=4x²+3x+6.
(b) Addition of algebraic expressions.
(c) Value: Economy is everywhere.

\(({1\over3}y^2-{4\over7}y^2+5)-({2\over7}y-{2\over3}y^3+2)={1\over3}y^2-{4\over7}y+5-{2\over7}y+{2\over3}y^2-2\)
\(=({1\over3}+{2\over3})y^3-({4\over7}+{2\over7})y+5-2\)
\(={3\over3}y^2-{6\over7}y+3\)
\(=y^2-{6\over7}y+3\)
\(=(2)^2-{6\over7}(2)+3\)
\(=4-{12\over7}+3\)
\(={28-12+21\over7} \)                        \([\because y=2]\)
\(={37\over7}\).

We have \(({x^3\over 5}-{3\over2}x^2+{2\over3}x+{1\over4})\)-\(({8\over 5}x^2-{2\over3}x^3+{3\over2}x-1)\)
= \({x^3\over 5}-{3\over2}x^2+{2\over3}x+{1\over4}\)-\(({8\over 5}x^2-{2\over3}x^3+{3\over2}x-1)\)
 =\(({1\over5}+{2\over3})x^3+(-{3\over2}-{8\over5})x^2+({2\over3}-{3\over2})x+({1\over4}+1)\)
\(={13x^3\over 15}-{31x^2\over10}-{5x\over 6}+{5\over4}\) 

Expression mn² + m²n + 8mn + 9
Term I :mn²
degree = 1 + 2 = 3
Term II :m²n
degree = 1 + 2 = 3
Term III : 8mn
degree = 0 + 1 + 1 = 2
Term IV: 9
degree = 0

We have, m - n, m + n
\(\therefore\) Sum=m-n+m+n =m+m-n+n
= m(l + 1)- n(-1 +1) = 2m-n xO= 2m
and difference = (m - n) - (m + n) = m - n - m - n
=m-m-n-n
= m(I-1) - n(l + 1) = m X 0 - 2n = - 2n

Like terms are 10pq, -7qp and 78qp (have same algebraic factors p, q); 7P and 2405 p (have same algebraic factor p); 8q and -100q(have same algebraic factor q); - P² q² and 12q²P² (have same algebraic factors p, p, q, 1); -23 and 41 (both are constants); -5p² and 701p² (have same algebraic factors p, p); 13p² q, qp² (have same algebraic factors p, p, q).

We know that, the numerical factor of a term is said to be the numerical coefficient or simply the coefficient of the same term. So, the terms and corresponding coefficients are given below:

2x² y, the only term is 2x² y

Here, first multiply the variable x with itself to obtain x². Then, multiply x² by the variable y to get x²y. Finally, multiply X² y by the constant 2 to obtain 2x² y.