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#### Page No 130:

(i) x increased by 12 is (x + 12).
(ii) y decreased by 7 is (- 7).
(iii) The difference of a and b, when a>b is (a - b).
(iv) The product of x and y is xy.
The sum of x and y is (x + y).
So, product of x and y added to their sum is xy + (x + y).
(v) One third of x is $\frac{x}{3}$.
The sum of a and b is (a + b).
$\therefore$ One-third of x multiplied by the sum of a and b =
(vi) 5 times x added to 7 times y =

(vii) Sum of x and the quotient of y by 5 is $\mathbf{x}\mathbf{+}\frac{\mathbf{y}}{\mathbf{5}}$.
(viii) x taken away from 4 is (4-x).
(ix) 2 less than the quotient of x by y is $\frac{\mathbf{x}}{\mathbf{y}}\mathbf{-}\mathbf{2}$.
(x) x multiplied by itself is $\mathrm{x}×\mathrm{x}={\mathbf{x}}^{\mathbf{2}}$.
(xi) Twice x increased by y is .
(xii)
Thrice x added to y squared is $\left(3×\mathrm{x}\right)+\left(\mathrm{y}×\mathrm{y}\right)=\mathbf{3}\mathbf{x}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}$.
(xiii) x minus twice y is $\mathrm{x}-\left(2×\mathrm{y}\right)=\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{y}$.
(xiv) x cubed less than y cubed is $\left(\mathrm{y}×\mathrm{y}×\mathrm{y}\right)-\left(\mathrm{x}×\mathrm{x}×\mathrm{x}\right)={\mathbf{y}}^{\mathbf{3}}\mathbf{-}{\mathbf{x}}^{\mathbf{3}}\mathbf{.}$
(xv) The quotient of x by 8 is multiplied by y is $\frac{\mathrm{x}}{8}×\mathrm{y}=\frac{\mathbf{xy}}{\mathbf{8}}$.

#### Page No 130:

Ranjit's score in English = 80 marks
Ranjit's score in Hindi = x marks
Total score in the two subjects = (Ranjit's score in English + Ranjit's score in Hindi)
∴ Total score in the two subjects = (80 + x) marks

#### Page No 130:

(i) b × b × b × ... 15 times = ${\mathbf{b}}^{\mathbf{15}}$
(ii) y × y × y × ... 20 times = ${\mathbf{y}}^{\mathbf{20}}$
(iii) 14 × a × a × a × a × b × b × b =
(iv) 6 × x × x × y × y = $6×\left(\mathrm{x}×x\right)×\left(y×y\right)=\mathbf{6}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}$
(v) 3 × z × z × z × y × y × x = $3×\left(\mathrm{z}×\mathrm{z}×z\right)×\left(\mathrm{y}×y\right)×x=\mathbf{3}{\mathbf{z}}^{\mathbf{3}}{\mathbf{y}}^{\mathbf{2}}\mathbf{x}$

#### Page No 130:

(i) ${\mathrm{x}}^{2}{\mathrm{y}}^{4}=\left(\mathrm{x}×\mathrm{x}\right)×\left(\mathrm{y}×\mathrm{y}×\mathrm{y}×\mathrm{y}\right)=\mathbf{x}\mathbf{×}\mathbf{x}\mathbf{×}\mathbf{y}\mathbf{×}\mathbf{y}\mathbf{×}\mathbf{y}\mathbf{×}\mathbf{y}$
(ii) $6{\mathrm{y}}^{5}=6×\left(\mathrm{y}×\mathrm{y}×\mathrm{y}×\mathrm{y}×\mathrm{y}\right)=\mathbf{6}\mathbf{×}\mathbf{y}\mathbf{×}\mathbf{y}\mathbf{×}\mathbf{y}\mathbf{×}\mathbf{y}\mathbf{×}\mathbf{y}$
(iii) $9{\mathrm{xy}}^{2}\mathrm{z}=9×\mathrm{x}×\left(\mathrm{y}×\mathrm{y}\right)×\mathrm{z}=\mathbf{9}\mathbf{×}\mathbf{x}\mathbf{×}\mathbf{y}\mathbf{×}\mathbf{y}\mathbf{×}\mathbf{z}$
(iv) $10{\mathrm{a}}^{3}{\mathrm{b}}^{3}{\mathrm{c}}^{3}=10×\left(\mathrm{a}×\mathrm{a}×\mathrm{a}\right)×\left(\mathrm{b}×\mathrm{b}×\mathrm{b}\right)×\left(\mathrm{c}×\mathrm{c}×\mathrm{c}\right)=\mathbf{10}\mathbf{×}\mathbf{a}\mathbf{×}\mathbf{a}\mathbf{×}\mathbf{a}\mathbf{×}\mathbf{b}\mathbf{×}\mathbf{b}\mathbf{×}\mathbf{b}\mathbf{×}\mathbf{c}\mathbf{×}\mathbf{c}\mathbf{×}\mathbf{c}$

#### Page No 132:

(i) a+b
Substituting a = 2 and b = 3 in the given expression:
2+3 = 5

(ii) ${\mathrm{a}}^{2}+\mathrm{ab}$
Substituting a = 2 and b = 3 in the given expression:
$\left(2{\right)}^{2}+\left(2×3\right)=4+6\phantom{\rule{0ex}{0ex}}=10$

(iii) $\mathrm{ab}-{\mathrm{a}}^{2}$
Substituting a = 2 and b = 3 in the given expression:
$\left(2×3\right)-\left(2{\right)}^{2}=6-4\phantom{\rule{0ex}{0ex}}=2$

(iv) 2a-3b
Substituting a = 2 and b = 3 in the given expression:
$\left(2×2\right)-\left(3×3\right)=4-9\phantom{\rule{0ex}{0ex}}=-5$

(v) $5{\mathrm{a}}^{2}-2\mathrm{ab}$
Substituting a=2 and b=3 in the given expression:
$5×\left(2{\right)}^{2}-2×2×3=5×4-12=20-12\phantom{\rule{0ex}{0ex}}=8$

(vi) ${\mathrm{a}}^{3}-{\mathrm{b}}^{3}$
Substituting a=2 and b=3 in the given expression:
${2}^{3}-{3}^{3}=2×2×2-3×3×3=8-27\phantom{\rule{0ex}{0ex}}=-19$

#### Page No 132:

(i) 3x-2y+4z
Substituting x = 1, y = 2 and z = 5 in the given expression:
$3×\left(1\right)-2×\left(2\right)+4×\left(5\right)=3-4+20\phantom{\rule{0ex}{0ex}}=19$

(ii)
Substituting x = 1, y = 2 and  z = 5 in the given expression:
${1}^{2}+{2}^{2}+{5}^{2}=\left(1×1\right)+\left(2×2\right)+\left(5×5\right)=1+4+25\phantom{\rule{0ex}{0ex}}=30$

(iii) $2{\mathrm{x}}^{2}-3{\mathrm{y}}^{2}+{\mathrm{z}}^{2}$
Substituting x = 1, y = 2 and z = 5 in the given expression:
$2×\left(1{\right)}^{2}-3×\left(2{\right)}^{2}+{5}^{2}=2×\left(1×1\right)-3×\left(2×2\right)+\left(5×5\right)=2-12+25\phantom{\rule{0ex}{0ex}}=15$

(iv) $\mathrm{xy}+\mathrm{yz}-\mathrm{zx}$
Substituting x = 1, y = 2 and z = 5 in the given expression:
$\left(1×2\right)+\left(2×5\right)-\left(5×1\right)=2+10-5\phantom{\rule{0ex}{0ex}}=7$

(v) $2{\mathrm{x}}^{2}\mathrm{y}-5\mathrm{yz}+{\mathrm{xy}}^{2}$
Substituting x = 1, y = 2 and z = 5 in the given expression:
$2×\left(1{\right)}^{2}×2-5×2×5+1×\left(2{\right)}^{2}=4-50+4\phantom{\rule{0ex}{0ex}}=-42$

(vi) ${x}^{3}-{y}^{3}-{z}^{3}\phantom{\rule{0ex}{0ex}}$
Substituting x = 1, y = 2 and z = 5 in the given expression:
${1}^{3}-{2}^{3}-{5}^{3}=\left(1×1×1\right)-\left(2×2×2\right)-\left(5×5×5\right)=1-8-125\phantom{\rule{0ex}{0ex}}=-132$

#### Page No 132:

(i) ${\mathrm{p}}^{2}+{\mathrm{q}}^{2}-{\mathrm{r}}^{2}$
Substituting p = -2, q = -1 and r = 3 in the given expression:
$\left(-2{\right)}^{2}+\left(-1{\right)}^{2}-\left(3{\right)}^{2}=\left(-2×-2\right)+\left(-1×-1\right)-\left(3×3\right)\phantom{\rule{0ex}{0ex}}⇒4+1-9=-4$

(ii) $2{\mathrm{p}}^{2}-{\mathrm{q}}^{2}+3{\mathrm{r}}^{2}$
Substituting p = -2, q = -1 and r = 3 in the given expression:

$2×\left(-2{\right)}^{2}-\left(-1{\right)}^{2}+3×\left(3{\right)}^{2}=2×\left(-2×-2\right)-\left(-1×-1\right)+3×\left(3×3\right)\phantom{\rule{0ex}{0ex}}⇒8-1+27=\mathbf{34}$

(iii) $\mathrm{p}-\mathrm{q}-\mathrm{r}$
Substituting p = -2, q = -1 and r = 3 in the given expression:
$\left(-2\right)-\left(-1\right)-\left(3\right)=-2+1-3\phantom{\rule{0ex}{0ex}}=-4$

(iv) ${\mathrm{p}}^{3}+{\mathrm{q}}^{3}+{\mathrm{r}}^{3}+3\mathrm{pqr}$
Substituting p = -2, q = -1 and r = 3 in the given expression:

$\left(-2{\right)}^{3}+\left(-1{\right)}^{3}+\left(3{\right)}^{3}+3×\left(-2×-1×3\right)\phantom{\rule{0ex}{0ex}}=\left(-2×-2×-2\right)+\left(-1×-1×-1\right)+\left(3×3×3\right)+3×\left(6\right)\phantom{\rule{0ex}{0ex}}=\left(-8\right)+\left(-1\right)+\left(27\right)+18\phantom{\rule{0ex}{0ex}}=36$

(v) $3{\mathrm{p}}^{2}\mathrm{q}+5{\mathrm{pq}}^{2}+2\mathrm{pqr}$
Substituting p = -2, q = -1 and r = 3 in the given expression:

$3×\left(-2{\right)}^{2}×\left(-1\right)+5×\left(-2\right)×\left(-1{\right)}^{2}+2×\left(-2×-1×3\right)\phantom{\rule{0ex}{0ex}}=3×\left(-2×-2\right)×\left(-1\right)+5×\left(-2\right)×\left(-1×-1\right)+2×\left(-2×-1×3\right)\phantom{\rule{0ex}{0ex}}=-12-10+12\phantom{\rule{0ex}{0ex}}=-10$

(vi) ${\mathrm{p}}^{4}+{\mathrm{q}}^{4}-{\mathrm{r}}^{4}$
Substituting p = -2, q = -1 and r = 3 in the given expression:
$\left(-2{\right)}^{4}+\left(-1{\right)}^{4}-\left(3{\right)}^{4}\phantom{\rule{0ex}{0ex}}=\left(-2×-2×-2×-2\right)+\left(-1×-1×-1×-1\right)-\left(3×3×3×3\right)\phantom{\rule{0ex}{0ex}}=16+1-81\phantom{\rule{0ex}{0ex}}=-64$

#### Page No 132:

(i) Coefficient of x in 13x is 13.
(ii) Coefficient of y in -5y is -5.
(iii) Coefficient of a in 6ab is 6b.
(iv) Coefficient of z in -7xz is -7x.
(v) Coefficient of p in -2pqr is -2qr.
(vi) Coefficient of y2 in 8xy2z is 8xz.
(vii) Coefficient of x3 in  x3 is 1.
(viii) Coefficient of x2 in -x2 is -1.

#### Page No 132:

(i) Numerical coefficient of ab is 1.
(ii) Numerical coefficient of -6bc is -6.
(iii) Numerical coefficient of 7xyz is 7.
(iv) Numerical coefficient of −2x3y2z is -2.

#### Page No 132:

A term of expression having no literal factors is called a constant term.
(i) In the expression 3x2 + 5x + 8, the constant term is 8.
(ii) In the expression 2x2 − 9, the constant term is -9.
(iii) In the expression  ${4y}^{2}-5y+\frac{3}{5}$, the constant term is $\frac{3}{5}$.
(iv) In the expression ${z}^{3}{-2z}^{2}+z-\frac{8}{3}$ , the constant term is $-\frac{8}{3}$.

#### Page No 132:

The expressions given in (i), (iii), (vi) and (viii) contain only one term. So, each one of them is monomial.
The expressions given in (ii) and (ix) contain two terms. So, both of them are binomial.
The expressions given in (iv) and (v) contain  three terms. So, both of them are trinomial.
The expression given in (vii) contains four terms. So, it does not represents any of the given types.

#### Page No 133:

(i) Expression 4x5 − 6y4 + 7x2y − 9 has four terms, namely 4x5 ,-6y4 , 7x2y and -9.
(ii) Expression 9x3 − 5z4 + 7z3 y − xyz has four terms, namely 9x3 , -5z4 , 7z3 y and -xyz.

#### Page No 133:

The terms that have same literals are called like terms.
(i)  a2 and 2a2 are like terms.
(ii)  are like terms.
(iii) −2xy2 and 5y2x are like terms.
(iv) ab2c , acb2 , b2ac and cab2 are like terms.

#### Page No 134:

(i) Required sum = 3x + 7x
= (3+7)x = 10x

(ii) Required sum = 7y +(−9y)
= (7-9)y = -2y

(iii) Required sum = 2xy +5xy + (−xy)
= (2+5-1)xy = 6xy

(iv) Required sum = 3x+2y

(v) Required sum = 2x2 + (− 3x2) + 7x2
=(2-3+7)x2 = 6x2

(vi)Required sum =  7xyz + (− 5xyz) + 9xyz + (−8xyz)
= (7-5+9-8)xyz = 3xyz

(vii) Required sum = 6a3 +(− 4a3) + 10a3 +( −8a3)
=(6-4+10-8)a3 = 4a3

(viii) Required sum = x2 − a2 + (−5x2 + 2a2) +( −4x2 + 4a2 )
Rearranging and collecting the like terms =  x2 -5x2 -4x2 -a2 + 2a2 +4a2
= (1-5-4)x2 +(-1+2+4)a2
= -8x2 + 5a2

(i)

(ii)

(iii)

(iv)

#### Page No 134:

(i) Sum of the given expressions
= (3a − 2b + 5c)+(2a + 5b − 7c)+ (− a − b + c)
Rearranging and collecting the like terms
= 3a+2a-a-2b+5b-b+5c-7c+c
= (3+2-1)a + (-2+5-1)b + (5-7+1)c
= 4a+2b-c

(ii) Sum of the given expressions
= (8a − 6ab + 5b) + (−6a − ab − 8b) + (−4a + 2ab + 3b)
Rearranging and collecting the like terms
=(8−6−4)a + (- 6 −1+2)ab + (5− 8+ 3)b
= -2a-5ab+0 = -2a - 5ab

(iii) Sum of the given expressions
= (2x3 − 3x2 + 7x − 8) + (−5x3 + 2x2 − 4x + 1) + ( 3 − 6x + 5x2 − x3 )
Rearranging and collecting the like terms
=2x3−5x3 − x3 − 3x2 + 2x2 + 5x2 +7x-4x-6x-8+1+3
= (2-5-1)x3 +(-3+2+5)x2+(7-4-6)x-4
= -4x3 +4x2-3x-4

(iv) Sum of the given expressions
= (2x2 − 8xy + 7y2 − 8xy2)+( 2xy2 + 6xy − y2 + 3x2)+( 4y2 − xy − x2 + xy2 )
Rearranging and collecting the like terms
= 2x2 +3x2 − x2   + 7y2 − y2 +4y2 − 8xy + 6xy − xy− 8xy2 +2xy2 + xy2
= (2 +3− 1)x2   + (7 − 1 +4)y2 + (-8 + 6 −1)xy + (− 8 +2 +1)xy2
= 4x2   + 10y2 − 3xy -5xy2

(v) Sum of the given expressions
= (x3 + y3 − z3 + 3xyz)+(− x3 + y3 + z3 − 6xyz)+(x3 − y3 − z3 − 8xyz)
Rearranging and collecting the like terms
= x3 -x3 + x3 + y3 + y3 − y3 -z3 + z3 − z3 + 3xyz-6xyz-8xyz
= (1-1+1)x3 + (1+1-1)y3 + (-1+1-1)z3 +(3-6-8)xyz
= x3 + y3 − z3 -11xyz

(vi) Sum of the given expressions
= (2 + xx2 + 6x3)+(−6 −2x + 4x2 −3x3)+( 2 + x2)+( 3 − x3 + 4x − 2x2 )
Rearranging and collecting the like terms
= 6x3 −3x3x3x2 +4x2+ x2− 2x2+ x −2x+ 4x+2-6+2+3
=  (6-3-1)x3+(-1+4+1-2)x2+(1-2+4)x+1
= 2x3+2x2+3x+1

#### Page No 135:

Change the sign of each term of the expression that is to be subtracted and then add.

(i) Term to be subtracted = 5x
Changing the sign of each term of the expression gives -5x.

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

(ii)  Term to be subtracted = -xy
Changing the sign of each term of the expression gives xy.

6xy+xy
= (6+1)xy
= 7xy

(iii) Term to be subtracted = 3a
Changing the sign of each term of the expression gives -3a.

5b+(-3a)
= 5b-3a

(iv) Term to be subtracted = -7x
Changing the sign of each term of the expression gives 7x.
9y+7x

(v) Term to be subtracted = 10x2
Changing the sign of each term of the expression gives -10x2.
−7x2 + (-10x2) = −7x2 −10x2
= (−7−10)x2
= −17x2

(vi) Term to be subtracted = a2 − b2
Changing the sign of each term of the expression gives -a2 + b2.
b2 − a2 + (-a2 + b2) = b2 − a2 -a2 + b2
= (1+1)b2  +(−1-1) a2
= 2b2 − 2a2

#### Page No 135:

Change the sign of each term of the expression that is to be subtracted and then add.

(i) Term to be subtracted = 5a + 7b − 2c
Changing the sign of each term of the expression gives -5a -7b + 2c.
(3a − 7b + 4c)+(-5a -7b + 2c ) = 3a − 7b + 4c-5a -7b + 2c
= (3-5)a+( − 7-7)b + (4+2)c
= -2a − 14b + 6c

(ii) Term to be subtracted = a − 2b − 3c
Changing the sign of each term of the expression gives -a +2b + 3c.
(−2a + 5b − 4c)+(-a +2b + 3c ) = −2a + 5b − 4c-a +2b + 3c
= (−2-1)a + (5+2)b +(−4+3)c
= −3a + 7b − c

(iii) Term to be subtracted = 5x2 − 3xy + y2
Changing the sign of each term of the expression gives  -5x2 + 3xy - y2.

(7x2 − 2xy − 4y2)+(-5x2 + 3xy - y2) = 7x2 − 2xy − 4y2-5x2 + 3xy - y2
= (7-5)x2 +(−2+3)xy +(−4-1)y2
= 2x2 +xy − 5y2

(iv) Term to be subtracted = 6x3 − 7x2 + 5x − 3
Changing the sign of each term of the expression gives  -6x3 + 7x2 - 5x + 3.
(4 − 5x + 6x2 − 8x3)+(-6x3 + 7x2 - 5x + 3) = 4 − 5x + 6x2 − 8x3-6x3 + 7x2 - 5x + 3
= (-8-6)x3 +(6+7)x2 +(-5- 5)x + 7
= -14x3 + 13x2 - 10x + 7

(v) Term to be subtracted = x3 + 2x2y + 6xy2 − y3
Changing the sign of each term of the expression gives  -x3 - 2x2y - 6xy2 + y3.
(y3 − 3xy2 − 4x2y)+(-x3 - 2x2y - 6xy2 + y3) = y3 − 3xy2 − 4x2y-x3 - 2x2y - 6xy2 + y3
= -x3 +(- 2-4)x2y +(-6-3)xy2 + (1+1)y3
= -x3 - 6x2y - 9xy2 + 2y3

(vi) Term to be subtracted = −11x2y2 + 7xy −6
Changing the sign of each term of the expression gives  11x2y2 -7xy +6.
(9x2y2 −6xy + 9)+(11x2y2 -7xy +6) = 9x2y2 −6xy + 9+11x2y2 -7xy +6
= (9+11)x2y2 (-7−6)xy + 15
= 20x2y2 −13xy +15

(vii) Term to be subtracted = −2a + b + 6d
Changing the sign of each term of the expression gives 2a-b-6d.
(5a − 2b -3c)+(2a-b-6d ) = 5a − 2b -3c +2a-b-6d
= (5+2)a+(− 2-1)b -3c -6d
= 7a − 3b-3c -6d

#### Page No 135:

(i) 2p3 − 3p2 + 4p − 5 − 6p3 + 2p2 − 8p − 2 + 6p + 8
Rearranging and collecting the like terms
= (2-6)p3 +(−3+2)p2 + (4-8+6)p − 5-2+8
= -4p3 −p2 +2p +1

(ii) 2x2 − xy + 6x − 4y + 5xy − 4x + 6x2 + 3y
Rearranging and collecting the like terms
= (2+6)x2 +(−1+5) xy + (6-4)x +(− 4+3)y
= 8x2 + 4xy + 2x − y

(iii) x4 − 6x3 + 2x − 7 + 7x3 − x + 5x2 + 2 − x4
Rearranging and collectingthe like terms
= (1-1)x4 +(− 6+7)x3 + 5x2 +(2-1)x-7+ 2
= 0 +  x3 + 5x2 +x-5
= x3 + 5x2 +x-5

#### Page No 135:

(3x2 − 5x + 2) + (−5x2 − 8x + 6)
Rearranging and collecting the like terms:
(3-5)x2 +(− 5-8)x + 2 +6
= -2x2 − 13x + 8

Subtract 4x2 − 9x + 7 from -2x2 − 13x + 8.

Change the sign of each term of the expression that is to be subtracted and then add.

Term to be subtracted = 4x2 − 9x + 7
Changing the sign of each term of the expression gives -4x2 + 9x - 7.
( -2x2 − 13x + 8 )+(-4x2 + 9x - 7 )    = -2x2 − 13x + 8 -4x2 + 9x - 7
= ( -2-4)x2 +(−13+9)x + 8 -7
= -6x2 − 4x + 1

#### Page No 135:

A = 7x2 + 5xy − 9y2
B = −4x2 + xy + 5y2
C = 4y2 − 3x2 − 6xy

Substituting the values of A, B and C in A+B+C:
= (7x2 + 5xy − 9y2)+(−4x2 + xy + 5y2)+(4y2 − 3x2 − 6xy)
= 7x2 + 5xy − 9y2−4x2 + xy + 5y2+4y2 − 3x2 − 6xy

Rearranging and collecting the like terms:
(7-4-3)x2 + (5+1-6)xy +(−9+5+4)y2
= (0)x2 + (0)xy + (0)y2
= 0
$⇒\mathrm{A}+\mathrm{B}+\mathrm{C}=0$

#### Page No 135:

Let the expression to be added be X.
(5x3 − 2x2 + 6x + 7)+X = (x3 + 3x2x + 1)
X = (x3 + 3x2x + 1) - (5x3 − 2x2 + 6x + 7)
Changing the sign of each term of the expression that is to be subtracted and then adding:
X = (x3 + 3x2x + 1) + (-5x3 + 2x2 - 6x - 7)
X = x3 + 3x2x + 1-5x3 + 2x2 - 6x - 7

Rearranging and collecting the like terms:
X = (1-5)x3 + (3+2)x2 +(−1-6) x + 1-7
X = -4x3 + 5x2 − 7x -6

So, -4x3 + 5x2 − 7x -6 must be added to 5x3 − 2x2 + 6x + 7 to get the sum as x3 + 3x2x + 1.

#### Page No 135:

P = a2 − b2 + 2ab
Q = a2 + 4b2 − 6ab
R = b2 + 6
S = a2 − 4ab
T = −2a2 + b2 − ab + a

Adding P, Q, R and S:
P+Q+R+S
= (a2 − b2 + 2ab)+(a2 + 4b2 − 6ab)+(b2 + 6)+(a2 − 4ab )
= a2 − b2 + 2ab+a2 + 4b2 − 6ab+b2 + 6+a2 − 4ab

Rearranging and collecting the like terms:
= (1+1+1)a2 +(−1+4+1) b2 + (2-6-4)ab+6
P+Q+R+S = 3a2 +4b2 - 8ab+6

To find P + Q + R + S − T, subtract T = (−2a2 + b2 − ab + a) from P+Q+R+S = (3a2 +4b2 - 8ab+6).

On changing the sign of each term of the expression that is to be subtracted and then adding:
Term to be subtracted = −2a2 + b2 − ab + a
Changing the sign of each term of the expression gives 2a2 - b2 + ab - a.
(3a2 +4b2 - 8ab+6)+(2a2 - b2 + ab - a) = 3a2 +4b2 - 8ab+6+2a2 - b2 + ab - a
= (3+2)a2 +(4-1) b2 +(-8+1) ab - a+6

P + Q + R + S − T = 5a2 +3b2 -7 ab - a+6

#### Page No 135:

Let the expression to be subtracted be X.
(a3 − 4a2 + 5a − 6)-X = (a2 − 2a + 1)
X = (a3 − 4a2 + 5a − 6)- (a2 − 2a + 1)
Since '-' sign precedes the parenthesis, we remove it and change the sign of each term within the parenthesis.
X = a3 − 4a2 + 5a − 6- a2 + 2a - 1
Rearranging and collecting the like terms:
X = a3 +(− 4-1)a2 + (5+2)a − 6 - 1
X = a3 −5a2 + 7a − 7
So, a3 −5a2 + 7a − 7 must be subtracted from a3 − 4a2 + 5a − 6 to obtain a2 − 2a + 1.

#### Page No 135:

To calculate how much is a + 2b − 3c greater than 2a − 3b + c, we have to subtract 2a − 3b + c from a + 2b − 3c.

Change the sign of each term of the expression that is to be subtracted and then add.

Term to be subtracted = 2a − 3b + c
Changing the sign of each term of the expression gives -2a + 3b - c.
(a + 2b − 3c )+(-2a + 3b - c )
= a + 2b − 3c -2a + 3b - c
= (1-2)a + (2+3)b +(− 3-1)c
= -a + 5b − 4c

#### Page No 135:

To calculate how much less than x − 2y + 3z is 2x − 4y − z, we have to subtract 2x − 4y − z from x − 2y + 3z.

Change the sign of each term of the expression that is to be subtracted and then add.

Term to be subtracted = 2x − 4y − z
Changing the sign of each term of the expression gives -2x + 4y + z.
(x − 2y + 3z)+(-2x + 4y + z )
= x − 2y + 3z-2x + 4y + z
= (1-2)x +(−2+4)y + (3+1)z
= -x + 2y + 4z

#### Page No 135:

To calculate how much does 3x2 − 5x + 6 exceed x3 − x2 + 4x − 1, we have to subtract x3 − x2 + 4x − 1 from 3x2 − 5x + 6.
Change the sign of each term of the expression that is to be subtracted and then add.

Term to be subtracted = x3 − x2 + 4x − 1
Changing the sign of each term of the expression gives -x3 + x2 - 4x + 1.
(3x2 − 5x + 6)+(-x3 + x2 - 4x + 1 )
= 3x2 − 5x + 6-x3 + x2 - 4x + 1
= -x3 + (3+1)x2 +(-5-4)x+6 + 1
= -x3 +4 x2 - 9x + 7

#### Page No 135:

Add 5x − 4y + 6z and −8x + y − 2z.

(5x − 4y + 6z )+(−8x + y − 2z)
= 5x − 4y + 6z −8x + y − 2z
= (5-8)x +(−4+1)y + (6-2)z
= -3x − 3y + 4z

Adding 12x − y + 3z and −3x + 5y − 8z:
(12x − y + 3z )+(−3x + 5y − 8z)
= 12x − y + 3z −3x + 5y − 8z
= (12-3)x +(−1+5)y + (3-8)z
= 9x +4y -5z

Subtract -3x − 3y + 4z from 9x +4y -5z.
Change the sign of each term of the expression that is to be subtracted and then add.

Term to be subtracted = -3x − 3y + 4z
Changing the sign of each term of the expression gives 3x + 3y - 4z.
(9x +4y -5z)+(3x + 3y - 4z )
= 9x +4y -5z+3x + 3y - 4z
= (9+3)x +(4+3)y + (-5-4)z
= 12x +7y -9z

#### Page No 135:

To calculate how much is 2x − 3y + 4z greater than 2x + 5y − 6z + 2, we have to subtract 2x + 5y − 6z + 2 from 2x − 3y + 4z.
Change the sign of each term of the expression that is to be subtracted and then add.

Term to be subtracted = 2x + 5y − 6z + 2
Changing the sign of each term of the expression gives -2x - 5y + 6z - 2.
(2x − 3y + 4z )+(-2x - 5y + 6z - 2 )
= 2x − 3y + 4z-2x - 5y + 6z - 2
= (2-2)x + (-3-5)y +(4+6)z-2
= 0-8y+10z-2
= -8y+10z-2

#### Page No 135:

To calculate how much does 1 exceed 2x-3y-4, we have to subtract 2x-3y-4 from 1.
Change the sign of each term of the expression to be subtracted and then add.

Term to be subtracted = 2x-3y-4
Changing the sign of each term of the expression gives -2x+3y+4.
(1)+(-2x+3y+4 )
= 1-2x+3y+4
= 5-2x+3y

#### Page No 136:

a - (b - 2a)
Here,  '-' sign precedes the parenthesis. So, we will remove it and change the sign of each term within the parenthesis.
=a - b + 2a
=3a - b

#### Page No 136:

4x − (3y − x + 2z)
Here, '−' sign precedes the parenthesis. So, we will remove it and change the sign of each term within the parenthesis.
= 4x − 3y + x − 2z
= 5x − 3y − 2z

#### Page No 136:

(a2 + b2 + 2ab) − (a2 + b2 − 2ab)
Here, '−' sign precedes the second parenthesis. So, we will remove it and change the sign of each term within the parenthesis.
a2 + b2 + 2ab − a2 − b2 +2ab

Rearranging and collecting the like terms:
a2 − a2 +b2 − b2 + 2ab + 2ab
=(1 − 1)a2 + (1− 1)b2 + (2 + 2)ab
=0 + 0 + 4ab
= 4ab

#### Page No 136:

−3(a + b) + 4(2a − 3b) − (2a − b)
Here, '−' sign precedes the first and the third parenthesis. So, we will remove them and change the sign of each term within the two parenthesis.
= −3a − 3b + (4$×$2a )−(4$×$3b) − 2a + b
= − 3a − 3b + 8a − 12b − 2a + b

Rearranging and collecting the like terms:
−3a + 8a − 2a − 3b − 12b + b
= (−3 + 8 − 2)a + (−3 − 12 + 1)b
= 3a −14b

#### Page No 136:

−4x2 + {(2x2 − 3) − (4 − 3x2)}

We will first remove the innermost grouping symbol (  ) and then {  }.

∴ −4x2 + {(2x2 − 3) − (4 − 3x2)}
= −4x2 + {2x2 − 3 − 4 + 3x2}
= −4x2 + {5x2 − 7}
= −4x2 + 5x2 − 7
= x2 − 7

#### Page No 136:

−2(x2y2 + xy) −3(x2 + y2xy)
Here a '−' sign precedes both the parenthesis. So, we will remove them and change the sign of each term within the two parenthesis.
= −2x2 +2y2 − 2xy −3x2 − 3y2 + 3xy
= (−2 − 3)x2 +(2 − 3)y2 + (− 2 + 3)xy
= −5x2y2 + xy

#### Page No 136:

a − [2b − {3a − (2b − 3c)}]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ a − [2b − {3a − (2b − 3c)}]
= a − [2b − {3a − 2b + 3c}]
= a − [2b − 3a + 2b − 3c]
= a − [4b − 3a − 3c]
= a − 4b + 3a + 3c
= 4a − 4b + 3c

#### Page No 136:

−x + [5y − {x − (5y − 2x)}]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ −x + [5y − {x − (5y − 2x)}]
= −x + [5y − {x − 5y + 2x}]
= −x + [5y − {3x − 5y}]
= −x + [5y − 3x + 5y]
= −x + [10y − 3x]
= −x + 10y − 3x
=  − 4x + 10y

#### Page No 137:

86 − [15x − 7(6x − 9) −2{10x − 5(2 − 3x)}]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ 86 − [15x − 7(6x − 9) −2{10x − 5(2 − 3x)}]
= 86 − [15x − 42x + 63 −2{10x − 10 + 15x}]
= 86 − [15x − 42x + 63 −2{25x − 10}]
= 86 − [15x − 42x + 63 −50x + 20]
= 86 − [− 77x + 83 ]
= 86 + 77x − 83
= 77x + 3

#### Page No 137:

12x − [3x3 + 5x2 − {7x2 − (4 − 3x − x3) + 6x3} − 3x]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ 12x − [3x3 + 5x2 − {7x2 − (4 − 3x − x3) + 6x3} − 3x]
= 12x − [3x3 + 5x2 − {7x2 − 4 + 3x + x3+ 6x3} − 3x]
= 12x − [3x3 + 5x2 − {7x2 − 4 + 3x + 7x3} − 3x]
= 12x − [3x3 + 5x2 − 7x2 + 4 − 3x − 7x3 − 3x]
= 12x − [ − 2x2 + 4 − 4x3 − 6x]
= 12x + 2x2 − 4 + 4x3 + 6x
= 4x3 + 2x2 +18x-4

#### Page No 137:

5a − [a2 − {2a(1 − a + 4a2) − 3a(a2 − 5a − 3)}] −8a
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ 5a − [a2 − {2a(1 − a + 4a2) − 3a(a2 − 5a − 3)}] −8a
= 5a − [a2 − {2a − 2a2 + 8a3 − 3a3 + 15a2 + 9a}] −8a
= 5a − [a2 − {5a3 + 13a2 + 11a}] − 8a
= 5a − [a2 − 5a3 − 13a2 −11a] − 8a
= 5a − [ − 5a3 − 12a2 − 11a] − 8a
= 5a + 5a3 + 12a2 + 11a − 8a
= 5a3 + 12a2 + 8a

#### Page No 137:

3 − [x − {2y − (5x + y − 3) + 2x2} − (x2 − 3y)]

We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ 3 − [x − {2y − (5x + y − 3) + 2x2} − (x2 − 3y)]
= 3 − [x − {2y − 5x - y + 3 + 2x2} − x2 + 3y]
= 3 − [x − {y − 5x + 3 + 2x2} − x2 + 3y]
= 3 − [x − y + 5x − 3 − 2x2 − x2 + 3y]
= 3 − [ 6x − 3 − 3x2 + 2y]
= 3 − 6x + 3 + 3x2 − 2y
= 3x2 − 2y − 6x+6

#### Page No 137:

xy − [yzzx − {yx − (3yxz) − (xyzy)}]

We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ xy − [yzzx − {yx − (3yxz) − (xyzy)}]
= xy − [yzzx − {yx − 3y + xzxy + zy}]
= xy − [yzzx − {− 3y + xz + zy}]  $\left(\because \mathrm{xy}=\mathrm{yx}\right)$
= xy − [yzzx + 3y - xz - zy
= xy − [− 2zx + 3y
= xy + 2zx − 3y

#### Page No 137:

2a − 3b − [3a − 2b − {a − c − (a − 2b)}]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ 2a − 3b − [3a − 2b − {a − c − (a − 2b)}]
= 2a − 3b − [3a − 2b − {a − c − a + 2b}]
= 2a − 3b − [3a − 2b − {− c  + 2b}]
= 2a − 3b − [3a − 2b + c  − 2b]
= 2a − 3b − [3a − 4b + c ]
= 2a − 3b − 3a + 4b − c
= − a + b − c

#### Page No 137:

-a − [a + {a + b − 2a − (a − 2b)} − b]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ −a − [a + {a + b − 2a − (a − 2b)} − b]
= −a − [a + {a + b − 2a − a + 2b} − b]
= −a − [a + {3b − 2a } − b]
= −a − [a + 3b − 2a  − b]
= −a − [2b − a ]
= −a − 2b + a
= −2b

#### Page No 137:

$2a-\left[4b-\left\{4a-\left(3b-\overline{2a+2b}\right)\right\}\right]$
We will first remove the innermost grouping symbol bar bracket. Next, we will remove (  ), followed by {  } and then [   ].

$2a-\left[4b-\left\{4a-\left(3b-\overline{2a+2b}\right)\right\}\right]$
= 2a-[4b-{4a-(3b-2a-2b)}]
= 2a-[4b-{4a-(b-2a)}]
= 2a-[4b-{4a-b+2a}]
=2a-[4b-{6a-b}]
= 2a-[4b-6a+b]
= 2a-[5b-6a]
= 2a-5b+6a
= 8a-5b

#### Page No 137:

5x − [4y − {7x − (3z − 2y) + 4z − 3(x + 3y − 2z)}]
We will first remove the innermost grouping symbol (  ), followed by {  } and then [   ].

∴ 5x − [4y − {7x − (3z − 2y) + 4z − 3(x + 3y − 2z)}]
= 5x − [4y − {7x − 3z + 2y + 4z − 3x − 9y + 6z}]
= 5x − [4y − {4x + 7z − 7y}]
= 5x − [4y − 4x − 7z + 7y]
= 5x − [11y − 4x − 7z ]
= 5x − 11y + 4x + 7z
= 9x − 11y + 7z

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