# NCTB Class 7 Math Chapter six Exercise 6.1 Solution

NCTB Class 7 Math Chapter six Exercise 6.1 Solution by Math Expert. Bangladesh Board Class 7 Math Solution Chapter 6.1 “Algebraic Fractions” Exercise 6.1 Solution.

 Board NCTB Class 7 Subject Math Chapter Six Chapter Name “Algebraic Fractions” Exercise 6.1 Solution

## Exercise:- 6.1

(1) Solution:- a2 b/a3 c

= b/ac

(2) Solution:- a2 b c/ab2 c

= a/b

(3) Solution:-  x3y3z3/x2y2z2

= x y z

(4) Solution:-  x2+x/x y +y

= x(x+1)/y(x+1)

= X/y

(5) Solution:-  4a2 b/6a3 b

= 4/6x a2 b/a3 b

= 2/3×1/a

= 2/3a

(6) Solution:- 2a-4ab/1-4b2

= 2a (1-2b)/(1)2– (2b)2

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

= 2a/(1+2b)

(7) Solution:- 2a+3b/4a2-9b2

= 2a+3b/(2a)2 – (3b)2

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

= 1/2a-3b

(8) Solution:- a2+4a+4

= a2+2a+2a+4/(a)2 – (2)2

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

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

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

(9) Solution:- x2-y2/(x +y)2

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

= X-y/x +y

(10) Solution:- x2+2x-15/x2 +9x+20

= x2+5x-3×15/x2+5x+4x+20

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

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

(11) Solution:- a/b c, d/a c

Denominator b c 8 ac, L.C.M= a b c

Therefore, a/ac = a x a/b c x a

= a2/a b c

And a/ac = a x b/a c x b

= a b/a b c

(12) Solution:- x/p q, y/p r

L.C.M of denominator = p q r

Therefore, x/p q = x X r/p q x r = x r/p q r

y/p r= y X x/p r x q = y q/p q r

(13) Solution:- 2x/3m, 3y/2n

L.C.M of denominator= 6mn

Therefore, 2x/3m = 2xX2n/3mx2n= 4xn/6n

= 3y/2n = 3yx3m/2mx3m= 9my/6mn

(14) Solution:- a/a-b, b/a +b

Therefore, L.C.M of denominator (a +b) (a-b)

Therefore, a/a-b= a x(a +b)/(a-b) (a +b) = a(a +b)/a2 – b2

=> b /a +b = b x(a-b)/(a +b) (a-b) = b(a-b)/a2 – b2

(15) Solution:- x2/a2-2ab, y2/a+2b

Therefore, L.C.M of denominator= (a+2b) (a2-2ab)

= (a+2b) (a-2b)a

Therefore, x2/a2-2ab= x2X(a2+2ab)/(a2-2ab) (a2+2ab)= x2 (a2+2ab)/a (a2-4b2)

=> y2/a+2b= y2xa (a-2b)/(a+2b) (a-2b)a = ay2 (a-2b)/a (a2-4b2)

(16) Solution:- 3/a2-4, 2/a (a+2)

L.C.M, of denomination, = (a2-4) a (a+2)

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

3/a2-4 = 3xa/(a+2) (a-2)a

= 3a/a(a2-4)

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

= 2 (a-2)/a(a2-4)

(17) Solution:- a/a2-9, b/a+3

Therefore, a2-9, (a)2 – (3)2 = (a+3) (a-3)

L.C.M of denominator= a (a+3) (a-3)

a/a2-9 = a. 1/(a+3) (a-3) = a/a2 – 9

= b/(a+3) = b(a-3)/(a+3) (a-3) = b(a-3)/a2 -9

(19) Solution:- a/a-b, b/a +b, c/a(a +b)

L.C.M, of denominator = a (a +b) (a-b)

a/ a-b = a x a(a +b) /(a-b) a(a +b)= a2(a +b)/a(a2-b2)

b /a +b= b x a (a-b)/(a +b) x a (a-b) = a b(a-b)/a (a2-b2)

c/a (a +b) = c x(a-b)/a (a +b) (a-b) = c(a-b)/a (a2-b2)

(20) Solution:- 2/x2– x- 2, 3/x2+x-6

= x2-x-2 = (x+1) (x-2)

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

Therefore, (x+1) (x-2) (x+3)

Therefore, 2/x2-x-2= 2x(x+3)/(x+1) (x-2) x (x+3)

3/x2+x-6 = 3x(x+1)/(x+3) (x-2) (x-1)

Updated: December 17, 2020 — 4:20 pm