NCTB Class 7 Math Chapter six Exercise 6.2 Solution

NCTB Class 7 Math Chapter six Exercise 6.2 Solution by Math Expert. Bangladesh Board Class 7 Math Solution Chapter 6.2“Algebraic Fractions” Exercise 6.2 Solution.

Board NCTB
Class 7
Subject Math
Chapter Six
Chapter Name “Algebraic Fractions”
Exercise 6.2Solution

Exercise:- 6.2

(1) Solution:- 2/3a, 3/5ab

=>2/3a= 2x5b/3ax5b = 10b/15ab

= 3/5ab = 3×3/5abx3 = 9/15ab

(2) Solution:- x/y z, y/z x

Therefore, x/y z= x Xx/y z x x = x2/x y z

y/z x = y x y/z x X y = y2/x y z

(3) Solution:- 1/a +b + 1/a-b

= a-b +a +b/(a-b) (a-b)

(c) = 2a/a2-b2

(4) Solution:- x/2+1= 3

=> x/2 = 3-1

=> x= 4

(5) Solution:- a/b = a x c/b x c = ac/b c

(6) Solution:- 4a2 b- 9b3/4a2 b +6ab2

= b (2a=3b) (2a-3b)/2ab (2a+3b)

= b(2a-3b)/2ab

(7) Solution:- a/x+ b/x – c/x

= a +b-c/x

(8) Solution:- (a) (x+2) (x-2)

(9) Solution:- x2+4x+4/x2-4

= x2+2. X.2+4/(x+2) (x-2)

= (x+2)2/(x+2) (x-2)

= (x+2)2/(x+2) (x-2)

= X+2/x-2

(10) Solution:-  3a/5 + 2b/5 = 3a+2b/5

 (11) Solution:- 1/5x+ 2/5x

= 1+2/5x

= 3/5x

(12) Solution:- x/2z + y/3b

= 3bx+2ay/6ab

(13) Solution:- 2a/x+1 + 2a/x-2

= 2a(x-2) + 2a(x+1)/(x+1) (x-2)

= 4ax-4a+2a/(x+) (x-2)

= 4ax-2a/(x+1) (x-2)

= 2a(x-1)/(x+1) (x-2)

(14) Solution:- a/a+2 + 2/a-2

= a (a-2) + 2(a+2)/(a+2) (a-2)

= a2-2a+2a+4/a2-4

= a2+4/a2 – 4

(15) Solution:- 3/x2-4x-5 + 4/x+1

= 3/x2-5x+x-5 + 4/x+1

= 3/x(x-5) (x+1) + 4/x+1

= 3+ 49x-5)/(x-1) (x-5)

= 3+ 4x-20/(x+1) (x-5)

= 4x-17/(x+1) (x-5)

(16) Solution:- 2a/7 – 4b/7

= 2a-4b/7

(17) Solution:- 2x/5a – 4y/5a

= 2x-4y/ 5a

(18) Solution:- a/8x- b/4y

= ay-2bx/8xy

(19) Solution:- 3/x+3- 2/x+2

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

= 3x+6 -2x-6/(x+3) (x+2)

= x/(x+3) (x+2)

(20) Solution:- p + q/p q – q + r/q r

= p r + q r-p q-p r/p q r

= q(r-p)/p q r

= (r-p)/p r

(22) Solution:- 5/a2+6a+5 = 1/(a-1)

= 5/(a-1) (a-b) + 1/(a-1)

= 5+11 (a-5)/(a-1) (a-5)

= 5+ a-5/(a-1) (a-5)

= a/(a-1) (a-5)

(23) Solution:- 1/x+2 – 1/x2-4

= 1/x+2 – 1/(x+2) (x-2)

= x-2 – 1/(x+2) (x-2)

= x-3/x2-4

(24) Solution:- a/3 + a/6 -3a/8

= 8a+4a-9a/24

= 3a/24

= a/8

(25) Solution:- a/b – 3a/2b+ 2a/3b

= 6a-9a+4a/6b

= a/6b

(26) Solution:- x/y z – y/z x +z/x y

= x2-y2+z2/x y z

(27) Solution:- x-y/x y + y-z/y z + z-x/z x

= z x-y z+ x y-x z +y z-x y/ x y z

= 2/ x y z

= 0

(28) Solution:- (a) x2– 3xy – 4y2

= x2– 4xy + x y – 4y2

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

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

(b) Solution:- x/x +y , x/x-4y

= L.C.M of denominator= (x+ y) (x-4y)

=> x/x +y = x X(x-4y)/ (x+ y) (x-4y) = x(x-4y)/ (x-y) (x-4y)

=> x/x-4y = x X(x +y) /(x +y) (x-4y) = x(x +y)/(x +y) (x-4y)

(c) Solution:- x/x +y + x/x-4y + y/x2-3xy-4y2

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

= x2– 4xy + x2+ x y +y/ (x +y) (x- 4y)

= 2x2 – 3x y +y/ (x +y) (x-4y)

(29) Solution:- A= 1/x2+3x , B= 2/x2+5x+6

C= 3/x2– x-12

(a) Solution:- x2=5x+6

= x2+3x+2x+6

= x(x+3) +2(x+3)

= (x+3) (x+2)

(b) Solution:- x2 + 3x = X(x+3)

X2+ 5x+6 = (x=3) (x+2)

X2– x-12 = x2-4x+3x-12

= x(x-4) +3(x-4)

= (x+3) (x-4)

Therefore, L.C.M of denominator= x(x+2) (x+3) (x-4)

Let, A= 1/x2+3x

= 1x (x+2) (x-2)/x((x+3) x (x+2) (c-4)

= (x+2) (x-4)/x(x+2) (x+3) (x-4)

B= 2/x2+5x+6

=2/(x+2) (x+3)

= 2x(x-4) x/(x+2) (x+3) X x(x-4)

= 2x(x-4)/x(x+2) (x+3) (x-4)

C= 3/x2-x-12

= 3/(x+3) (x-4)

= 3Xx(x+2)/(x+3) (x-4) Xx(x+2)

= 3x(x+2)/x(x+2) (x+3) (x-4)

(c) Solution:-A+B-c

= (x+2) (x-4)/x(x+2) (x+3) (x-4) + 2x(x-4)/x(x+2) (x+3) (x-4) -3x(x+2) / x(x+2) (x+3) (x-4)

= x2 -4x+2x-8 +2×2-8+2x2-8x-3x2-6x/x(x+2) (x+3) (x-4)

= 3x2-3x2-18x+2x-8/x(x+2) (x+3) (x-4)

= -8-16x/x(x+2) (x+3) (x-4)

Updated: December 18, 2020 — 8:50 am

Leave a Reply

Your email address will not be published. Required fields are marked *