Algebraic Expressions ICSE Class-8th Concise Selina Mathematics Solutions Chapter-11 . We provide step by step Solutions of Exercise / lesson-11 Algebraic Expressions for ICSE Class-8 Concise Selina Mathematics. Our Solutions contain all type Questions with Exe-11 A, Exe-11 B, Exe-11 C, Exe-11 D and Exe-11 E to develop skill and confidence . Visit official Website CISCE for detail information about ICSE Board Class-8.
Algebraic Expressions ICSE Class-8th Concise Selina Mathematics Solutions Chapter-11
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Exercise – 11 A Algebraic Expressions for ICSE Class-8th Mathematics
Question 1 :-
Separate the constants and variables from the following :
Answer :-
Question 2 :-
Write the number of terms in each of the following polynomials.
(i) 5x2 + 3 x ax
(ii) ax ÷ 4 – 7
(iii) ax – by + y x z
(iv) 23 + a x b ÷ 2.
Answer :-
Question 3 :-
Separate monomials, binomials, trinomials and polynomials from the following algebraic expressions :
Answer :-
Question 4 :-
Write the degree of each polynomial given below :
Answer :-
Question 5 :-
Write the coefficient of :
(i) ab in 7abx ,
(ii) 7a in 7abx ;
(iii) 5x2 in 5x2 – 5x ;
(iv) 8 in a2 – 8ax + a ;
(v) 4xy in x2 – 4xy + y2.
Answer :-
Question 6 :-
In 
Answer :-
Question 7 :-
In each polynomial, given below, separate the like terms :
(i) 3xy, − 4yx2, 2xy2, 2.5x2y, −8yx, −3.2y2x and x2y
(ii) y2z3, xy2z3, −5x2yz, −4y2z3, −8xz3y2, 3x2yz and 2z3y2
Answer :-
(i) Like terms are 3xy, −8yx; −4yx2, 2.5x2y, and x2y; 2xy2 and −3.2y2x
(ii) Like terms are y2z3, −4y2z3, and 2z3y2; xy2z3 and −8xz3y2: −5x2yz and 3x2yz
Algebraic Expressions ICSE Class-8th Mathematics Exe – 11 B
Question 1 :-
Evaluate :
Answer :-
Question 2 :-
Add :
Answer :-
Question 3 :-
Find the total savings of a boy who saves ₹ (4x – 6y) ; ₹ (6x + 2y) ; ₹ (4y – x) and ₹ (y – 2x) for four consecutive weeks.
Answer :-
Question 4 :-
Subtract :
Answer :-
Question 5 :-
(i) Take away – 3x3 + 4x2 – 5x + 6 from 3x3 – 4x2 + 5x – 6
(ii) Take m2 + m + 4 from -m2 + 3m + 6 and the result from m2 + m + 1.
Answer :-
Question 6 :-
Subtract the sum of 5y2 + y – 3 and y2 – 3y + 7 from 6y2 + y – 2.
Answer :-
Question 7 :-
What must be added to x4 – x3 + x2 + x + 3 to obtain x4 + x2 – 1 ?
Answer :-
x4 + x2 – 1
+x4 – x3 + x2 + x + 3
– + – – –
x3 – x – 4
Question 8 :-
(i) How much more than 2x2 + 4xy + 2y2 is 5x2 + 10xy – y2 ?
(ii) How much less 2a2 + 1 is than 3a2 – 6 ?
Answer :-
(i)
5x2 + 10xy – y2
+2x2 + 4xy + 2y2
– – –
3x2 + 6xy – 3y2
(ii)
3a2 – 6
+ 2a2 + 1
– –
a2 – 7
Question 9 :-
If x = 6a + 86 + 9c ; y = 2b – 3a – 6c and z = c – b + 3a ; find
(i) x + y + z
(ii) x – y + z
(iii) 2x – y – 3z
(iv) 3y – 2z – 5x
Answer :-
(i) x = 6a + 8b + 9c
y = − 3a + 2b − 6c
z = +3a − b + c
Adding x + y + z = 6a + 9b + 4c
(ii) x − y + z = (6a + 8b + 9c) − (2b − 3a − 6c) + (c − b + 3a)
= 6a + 8b + 9c − 2b + 3a + 6c + c − b + 3a
= 6a + 3a + 3a + 8b − 2b − b + 9c + 6c + c
= 12a + 5b + 16c
(iii) 2x − y − 3z = 2(6a + 8b + 9c) − (2b − 3a − 6c) − 3(c −b + 3a)
= 12a + 16b + 18c − 2b + 3a + 6c − 3c +3b − 9a
= 12a + 3a − 9a + 16b + 3b − 2b + 18c + 6c −3c
= 6a + 17b + 21c
(iv) 3y − 2z − 5x = 3(2b − 3a − 6c) − 2(c − b + 3a) − 5(6a + 8b + 9c)
= 6b − 9a − 18c − 2c + 2b − 6a − 30a − 40b − 45c
= − 9a − 6a − 30a + 6b + 2b − 40b − 18c − 2c − 45c
= − 45a − 32b − 65c
Question 10 :-
The sides of a triangle are x2 – 3xy + 8, 4x2 + 5xy – 3 and 6 – 3x2 + 4xy. Find its perimeter.
Answer :-
Required perimeter = Sum of three sides
= x2 − 3xy + 8 + 4x2 + 5xy − 3 + 6 − 3x2 + 4xy
= x2 + 4x2 − 3x2 − 3xy + 5xy + 4xy + 8 − 3 + 6
= 2x2 + 6xy + 11
Question 11 :-
The perimeter of a triangle is 8y2 – 9y + 4 and its two sides are 3y2 – 5y and 4y2 + 12. Find its third side.
Answer :-
Perimeter of the triangle = Sum of three sides
= 8y2 – 9y + 4
Sum of two sides = 3y2 – 5y + 4y2 + 12
= 7y2 − 5y + 12
∴ (8y2 – 9y + 4) – (7y2 – 5y + 12)
= 8y2 – 9y + 4 – 7y2 + 5y – 12
= y2 – 4y – 8
Hence third side = y2 – 4y – 8
Question 12 :-
The two adjacent sides of a rectangle are 2x2 – 5xy + 3z2 and 4xy – x2 – z2. Find its perimeter.
Answer :-
Adjacent sides of a rectangle are 2x2 – 5xy + 3z2 and 4xy – x2 – z2
∴ Perimeter = 2(2x2 – 5xy + 3z2 + 4xy – x2 – z2)
= 4x2 – 10xy + 6z2 + 8xy – 2x2 – 2z2
= 2x2 – 2xy + 4z2
Question 13 :-
What must be subtracted from 19x4 + 2x3 + 30x – 37 to get 8x4 + 22x3 – 7x – 60 ?
Answer :-
The required result is
(22x – 20y – 13z + 26) – (15x – 18y + 19z)
= 22x – 20y – 13z + 26 – 15x + 18y – 19z
= 7x – 2y – 32z + 26
Question 14 :-
How much smaller is 15x – 18y + 19z than 22x – 20y – 13z + 26 ?
Answer :-
The required result is
(22x – 20y – 13z + 26) – (15x – 18y + 19z)
= 22x – 20y – 13z + 26 – 15x + 18y – 19z
= 7x – 2y – 32z + 26
Question 15 :-
How much bigger is 15x2y2 – 18xy2 – 10x2y than -5x2 + 6x2y – 7xy ?
Answer :-
The required result,
(5x2y2 – 18xy2 – 10x2y) – (–5x2 + 6x2y – 7xy)
= 5x2y2 – 18xy2 – 10x2y + 5x2 + 6x2y – 7xy
= 5x2y2 – 18xy2 – 16x2y + 5x2 + 7xy
ICSE Class-8th Concise Selina Maths Exe – 11 C
Question 1 :-
Multiply :
Answer :-
Question 2 :-
Multiply :
Answer :-
(i)
5x2 − 8xy + 6y2 − 3 × −3xy
= −15x3y3 + 24x2y2 − 18xy3 + 9xy
(ii)
3-(2/3)xy + (5/7)xy2 – (16/21)x2y × (-21x2y2)
= −63x2y2 + 14x3y3 − 15x3y4 + 16x4y3
(iii)
6x3 − 5x + 10
× 4 − 3x2
24x3 − 20x + 40
− 18x5 + 15x3 − 30x2
− 18x5 + 39x3 − 30x2 − 20x + 40
(iv)
2y − 4y3 + 6y5
× y2 + y − 3
2y3 − 4y5 + 6y7
+ 2y2 − 4y4 + 6y6
− 6y + 12y3 − 18y5
6y7 + 6y6 −(4+18)y5 − 4y4 + (2+12)y3 + 2y2 − 6y
= 6y7 + 6y6 − 22y5 − 4y4 + 14y3 + 2y2 − 6y
(v)
5p2 + 25pq + 4q2
× 2p2 − 2pq + 3q2
10p4 + 50p3q + 8p2q2
−10p3q − 50p2q2 − 8pq3
+15p2q2 + 75pq3 + 12q4
10p4 + 40p3q − 27p2q2 + 67pq3 + 12q4
Question 3 :-
Simplify :
(i) (7x – 8) (3x + 2)
(ii) (px – q) (px + q)
(iii) (5a + 5b – c) (2b – 3c)
(iv) (4x – 5y) (5x – 4y)
(v) (3y + 4z) (3y – 4z) + (2y + 7z) (y + z)
Answer :-
(i)
(7x – 8) (3x + 2) = 7x(3x + 2) −8(3x+ 2)
= 21x2 + 14x – 24x – 16
= 21x2 – 10x – 16
(ii)
(px – q) (px + q) = px(px + q) – q(px + q)
= p2x2 + pxq – pqx – q2
= p2x2 – q2
(iii)
(5a + 5b – c) (2b – 3c)
= 5a(2b – 3c) + 5b(2b – 3c) – c(2b – 3c)
= 10ab – 15ac + 10b2 – 15ac – 2bc + 3c2
= 10ab + 10b2 – 17bc – 15ac + 3c2
(iv)
(4x – 5y) (5x – 4y)
= 4x(5x – 4y) – 5y(5x – 4y)
= 20x2 – 16xy – 25xy + 20y2
= 20x2 – 41xy + 20y2
(v)
(3y + 4z) (3y – 4z) + (2y + 7z) (y + z)
= 3y (3y – 4z) + 4z (3y – 4z) + 2y (y + z) (y + z)
= 9y2 – 12yz + 12yz – 16z2 + 2y2 + 2yz + 7yz + 7z2
= (9+2)y2 + (–12 + 12 + 2 + 7)yz + (–16 + 7)z2
= 11y2 + 9yz – 9z2
Question 4 :-
The adjacent sides of a rectangle are x2 – 4xy + 7y2 and x3 – 5xy2. Find its area.
Answer :-
Reqd. area = (x2 – 4xy + 7y2) (x3 – 5xy2)
= x2 (x3 – 5xy2) – 4xy (x3 – 5xy2) + 7y2 (x3 – 5xy2)
= x5 – 5x3y2 – 4x4y + 20x2y3 + 7x3y2 – 35xy4
= x5 + 2x3y2 – 4x4y + 20x2y3 – 35xy4
= (x5 – 4x4y + 2x3y2 + 20x2y3 – 35xy4) sq. unit.
Question 5 :-
The base and the altitude of a triangle are (3x – 4y) and (6x + 5y) respectively. Find its area.
Answer :-
Reqd. Area = 1/2 (base) x (altitude)
=1/2 (3x − 4y) (6x + 5y)
=1/2 (18x2 + 15xy − 24xy − 20y2)
=1/2 (18x2 − 9xy − 20y2) sq. unit.
Question 6 :-
Multiply -4xy3 and 6x2y and verify your result for x = 2 and y= 1.
Answer :-
(−4xy3) × (6x2y) = (−4 × 6) (x x x2) (y3 x y)
= −24x3y4
For x = 2 and y = 1
(−4xy3) × (6x2y) = (−4 × 2 × 13) × (6 × 22 × 1)
= (−8) × 24 = −192
And, −24x3y4 = −24 × 23 × 14
= −24 × 8 × 1 = −192
∴ For x = 2 and y = 1, it is verified that
(−4xy3) × (6x2y) = −24x3y4
Question 7 :-
Find the value of (3x3) x (-5xy2) x (2x2yz3) for x = 1, y = 2 and z = 3.
Answer :-
For x = 1, y = 2 and z = 3
(3x3) × (−5xy2) × (2x2yz3)
(3 × 13) × (−5 × 1 × 22) × (2 × 12 × 2 × 33)
3 × (−5 × 4) × (2 × 1 × 2 × 27)
3 × (−20) × 108
= −6480
Question 8 :-
Evaluate (3x4y2) (2x2y3) for x = 1 and y = 2.
Answer :-
(3x4y2) (2x2y3)
(3 × 14 × 22) × (2 × 12 × 23)
(3 × 1 × 4) × (2 × 1 × 8)
= 12 × 16
= 192
Question 9 :-
Evaluate (x5) × (3x2) × (-2x) for x = 1.
Answer :-
For x = 1
(x5) × (3x2) × (−2x)
(15) × (3 × 12) × (−2 × 1)
1 × 3 × (−2)
= −6
Question 10 :-
If x = 2 and y = 1; find the value of (-4x2y3) × (-5x2y5).
Answer :-
For x = 2 and y = 1
(−4x2y3) × (−5x2y5)
(−4 × 22 × 13) × (−5 × 22 × 15)
(−4 × 4 × 1) × (−5 −4 × 1)
−16 × −20
= 320
Question 11 :-
Evaluate:
(i) (3x – 2)(x + 5) for x = 2.
(ii) (2x – 5y)(2x + 3y) for x = 2 and y = 3.
(iii) xz (x2 + y2) for x = 2, y = 1 and z= 1.
Answer :-
(i)
For x = 2
(3x – 2) (x + 5)
(3 × 2 − 2) (2 + 5)
(6 − 2) × 7
4 × 7
= 28
(ii)
(2x – 5y) (2x + 3y)
= 2x × 2x − 5y × 2x + 2x × 3y − 5y × 3y
= 4x2 − 10xy + 6xy − 15y2
Hence, x = 2 and y = 3
4x2 − 10xy + 6xy − 15y2
= 4(2)2 − 10(2)(3) + 6(2)(3) − 15(3)2
= 16 − 60 + 36 − 135
= 52 − 195
= −143
(iii)
For x = 2, y = 1 and z = 1
xz (x2 + y2)
2(1) × (22 + 12)
2 × (4 + 1)
2 × 5
= 10
Question 12 :-
Evaluate:
(i) x(x – 5) + 2 for x = 1.
(ii) xy2(x – 5y) + 1 for x = 2 and y = 1.
(iii) 2x(3x – 5) – 5(x – 2) – 18 for x = 2.
Answer :-
(i)
For x = 1
x(x – 5) + 2
1(1 − 5) + 2
−4 + 2
= −2
(ii)
For x = 2 and y = 1
xy2(x – 5y) + 1
2 × 12 (2 − 5 × 1) + 1
2 × (2 − 5) + 1
2 × (−3) + 1
= −5
(iii)
For x = 2
2x(3x – 5) – 5(x – 2) – 18
2 × 2(3 × 2 − 5) − 5(2 − 2) − 18
4(6 − 5) − 5 × 0 − 18
4 − 18
= − 14
Question 13 :-
Multiply and then verify :
-3x2y2 and (x – 2y) for x = 1 and y = 2.
Answer :-
Question 14 :-
Multiply:
(i) 2x2 – 4x + 5 by x2 + 3x – 7
(ii) (ab – 1)(3 – 2ab)
Answer :-
Question 15 :-
Simplify : (5 – x)(6 – 5x)(2 -x).
Answer :-
Algebraic Expressions Solved Questions of Exe-11 D for ICSE Class-8th Concise Selina Mathematics
Question 1 :-
Divide :
Answer :-
(iv) 
(v)
= −7a4−2 . b5−4 . c6−3
= −7a2bc3
(vi) 
(ix)
a+4)a2+7a+12¯(a+3
a2 +4a
− −
3a + 12
3a + 12
− −
×
∴ a = 3
(x)
x-6)x2+3x-54¯(x+9
x2 -6x
− +
9x − 54
+ 9x − 54
− +
×
∴ x + 9
(xi)
3x+4y)12x2+7xy-12y2¯(4x-3y
12x2 +16xy
− −
−9xy − 12y2
−9xy − 12
+ +
×
∴ 4x − 3y
(xii)
x2-2 ) x6-8 ¯(x4+2x2+4
x6 -2×4
− +
2x4 − 8
2x4 − 4x2
− +
4x2 − 8
4x2 − 8
− +
×
∴ x4 + 2x2 + 4
(xiii)
2x3-x-6)6x3-13x2-13x+30¯(3x-5
6x3-3x2-18x
− + +
−10x2 + 5x + 30
−10x2 + 5x + 30
+ − −
×
∴ 3x − 5
(xiv)
2a+3b+5c)4a2+12ab+9b2-25c2¯(2a+3b-5c
4a2 + 6ab + 10ca
− − −
6ab + 9b2 − 25c2 − 10ca
6ab + 9b2 +15bc
− − −
−10ca −25c2 −15bc
−10ca −25c2 −15bc
+ + +
×
∴ 2a + 3b − 5c
(xv)
-x3+x+4)+ x6-2x-8 x3+ x2+8x+16¯(- x3+x+4
+ x6 −x4 − 4x3
− + +
−x4 −4x3 + x2 + 8x +16
−x4 + x2 + 4x
+ − −
−4x3 + 4x + 16
−4x3 + 4x + 16
+ − −
×
∴ −x3 + x + 4
Question 2 :-
.
Find the quotient and the remainder (if any) when :
(i) a3 − 5a2 + 8a + 15 is divided by a + 1. verify your answer.
(ii) 3x4 + 6x3 − 6x2 + 2x − 7 is divided by x − 3. verify your answer.
(iii) 6×2 + x − 15 is divided by 3x +5. verify your answer.
Answer :-
(i)
a+1) a3-5 a2+8a+15¯( a2-6a+14
a3 + a2
− −
−6a2 + 8a + 15
−6a2 − 6a
+ +
14a + 15
14a + 14
1
∴ Quotient = a2 − 6a + 14 and reminder = 1
Verification :
Dividiend = Quotient × Divisor + Reminder
= (a2 − 6a + 14) × (a + 1) + 1
= a3 − 6a2 + 14a + a2 − 6a + 14 + 1
= a3 − 5a2 + 8a + 15 which is given
(ii)
-x-3)3x4+6x3-6x2+2x-7¯(3x3+15x2+39x+119
3x4 − 9x3
− +
15x3 − 6x2 + 2x − 7
15x3 − 45x2
− +
39x2 + 2x − 7
39x2 − 117x
− +
119x − 7
119x − 357
− +
350
∴ Quotient = 3x3 + 15x2 + 39x + 119 and reminder = 350
Verification :
Dividend = Quotient × Divisor + Reminder
= (3x3 + 15x2 + 39x + 119) (x − 3) + 350
= 3x4 + 15x3 + 39x2 + 119x − 9x3 − 45x2 − 117x − 357 + 350
= 3x4 + 6x3 − 6x2 + 2x − 7 which is given
(iii)
3x+5)6x2 +x−15¯(2x−3
6x2 + 10x
− −
−9x − 15
−9x − 15
+ +
×
∴ Quotient = 2x − 3 and remainder = 0
Verification :
Dividend = Quotient × Divisor + Reminder
= (2x − 3) (3x + 5) + 0
= 6x2 + 10x − 9x − 15 + 0
= 6x2 − x − 15 which is given
Question 3 :-
The area of a rectangle is x3 – 8x2 + 7 and one of its sides is x – 1. Find the length of the adjacent side.
Answer :-
Area = x3 − 8x + 7
One side = x − 1
∴ Adjacent side = (x3 − 8x + 7) ÷ (x − 1)
x-1)x3-8x+7¯(x-7x-7
x3 − x2
− +
− 7x2 + 7x
− 7x2 + 7x
+ −
− 7x + 7
− 7x + 7
+ −
×
∴ Other side = x2 − 7x − 7
Question 4 :-
The product of two numbers is 16x4 – 1. If one number is 2x – 1, find the other.
Answer :-
Product of two numbers = 16x4 − 1
One number = 2x − 1
Then the second number =16×4-12x-1
= 8x3 + 4x2 + 2x + 1
2x-1)16x4 -1¯(8x3 +4x2 +2x+1
16x4 − 8x3
− +
8x3
8x3 − 4x2
− +
4x2
4x2 − 2x
− +
2x − 1
2x − 1
− +
×
∴ Other side = 8x3 + 4x2 + 2x + 1
Question 5 :-
Divide x6 – y6 by the product of x2 + xy + y2 and x – y.
Answer :-
Product of (x2 + xy + y2) and (x − y)
= (x − y) (x2 + xy + y2)
= x(x2 + xy + y2) − y(x2 + xy + y2)
= x3 + x2y + xy2 − x2y − xy2 − y3
= x3 − y3
Now, (x6 − y6) ÷ (x3 − y3)
= x3 + y3
x3-y3)x6 -x6¯(x3+y3
x6 − x3y3
− +
x3y3 − y6
x3y3 − y6
− +
×
Exercise – 11 E Algebraic Expressions
Question 1 :-
Simplify :
a2 − 2a + {5a2 − (3a + 4a2)}
Answer :-
= a2 − 2a + {5a2 − 3a + 4a2}
= a2 − 2a + {9a − 3a}
= a2 − 2a + 9a2 − 3a
= 10a2 − 5a
Question 2 :-
Answer :-
= x − y − {x − y − (x + y) − x + y}
= x − y − {x − y − x − y − x + y}
= x − y − x + y + x + y + x − y
= 2x
Question 3 :-
3 (1 − x2) − 2{x2 − (3 − 2x2)}
Answer :-
−3 (1 − x2) − 2 {x2 − (3 − 2x2)}
= −3 + 3x2 − 2 {x2 − 3 + 2x2}
= −3 + 3x2 − 2{3x2 − 3}
= −3 + 3x2 − 6x2 + 6
= 3 − 3x2
Question 4 :-
Answer :-
= 2{m − 3 (n + m − 2n)}
= 2{m − 3 (m − 3 (m − n)}
= 2{m − 3m + 3n}
= 2{3n − 2m}
= 6n − 4m
Question 5 :-
Answer :-
= 3x − [3x − {3x − (3x − 3x + y)}]
= 3x − [3x − {3x − y}]
= 3x − [3x − 3x + y]
= 3x − y
Question 6 :-
Answer :-
= p2x − 2 {px − 3x (x2 − 3a + x2)}
= p2x − 2{px − 3x (2x2 − 3a)}
= p2x − 2{px − 6x3 + 9ax}
= p2x − 2px + 12x3 − 18ax
Question 7 :-
Answer :-
= 2[6 + 4 {m − 6 (7 − n − p) + q }]
= 2[6 + 4 {m − 42 + 6n + 6p + q}]
= 2[6 + 4m − 168 + 24n + 24p + 4q]
= 2[4m + 24n + 24p + 4q − 162]
= 8m + 48n + 48p + 8q − 324
Question 8 :-
Answer :-
= a − [a − b − a − {a − (a − b + a)}]
= a − [− b − {a − a + b − a}]
= a − [− b − b + a]
= a + b + b − a
= 2b
Question 9 :-
Answer :-
= 3x − [4x − 3x + 5 − 3 {2x − (3x − 2x + 3y)}]
= 3x − [4x − 3x + 5y − 3 {2x − (x + 3y)}]
= 3x − [4x − 3x + 5y − 3 {2x − x − 3y}]
= 3x − [x + 5y − 6x + 3x + 9y]
= 3x − [−2x + 14y]
= 3x + 2x − 14y
= 5x − 14y
Question 10 :-
Answer :-
a5 ÷ a3 + 3a × 2a
= a5−3 + 3a × 2a
= a2 + 6a2
= 7a2
Question 11 :-
x5 ÷ (x2 × y2) × y3
Answer :-
x5 ÷ (x2 × y2) × y3
=( x5 x2 /y2) ×y3
= x5−2 − y3−2
= x3y
Question 12 :-
(x5 ÷ x2) × y2 × y3
Answer :-
(x5 ÷ x2) × y2 × y3
= x5−2 × y2+3
= x3y5
Question 13 :-
(y3 − 5y2) ÷ y × (y − 1)
Answer :-
(y3 − 5y2) ÷ y × (y − 1)
=(y3-5y2)/y×y-1
= (y2 − 5y) × (y − 1)
= y2 (y − 1) − 5y (y − 1)
= y3 − y2 − 5y2 + 5y
= y3 − 6y2 + 5y
Question 14 :-
Answer :-
= 3a × [2b − 6 {a − 7a + 3b}]
= 3a × [2b − 6 {−6a + 3b}]
= 3a × [2b + 36a − 18b]
= 3a × [36a − 16b]
= 108a2 − 48ab
Question 15 :-
7x + 4 {x2 ÷ (5x ÷ 10)} − 3 {2 − x3 ÷ (3x2 ÷ x)}
Answer :-
7x + 4 {x2 ÷ (5x ÷ 10)} − 3 {2 − x3 ÷ (3x2 ÷ x)}
= 7x + 8x − 6 +x2
= x2 + 15x − 6
— End of Algebraic Expressions Solutions :–
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