NCTB Class 6 Math Chapter Four Exercise 4.3 Solution by Math Expert. Bangladesh Board Class 6 Math Solution Chapter 4 “Algebraic Expressions’’ Exercise 4.3 Solution.
(1) Which one of the following is the coefficient of x in 5x+3y?
In 5x+3y, the coefficient of x is (d) 5
(2)Which one of the following is the sum of three times of x and two times of y?
(3) Which one is the index of x in 7x3 X x2?
In 7x3 X x2 = 7x2
Therefore, index of x is (b) 5
(4) Which one of the following indicates a pair of similar terms?
Pair of similar terms (c) 3x2, -7x2
(5) If m=-6 in expression m2-7, what is the value of the expression?
M = 6;
= (-6)2 -7
(6) If you Subtract b-a, what is difference?
= a-b-b +a
(7) What is the sum of three expressions x2+3, x2-2, -2x2+1?
(8) In the expressions 5x4
(b) (i) and (iii) correct.
(9) In x and y variables—
(i) (a) (i) and (ii) correct.
(10) If x=2 and y=-3, what is the value of the first expression?
Here, x=2, y= -3;
= (x+y) (x-y)
= (2-3) (2+3)
= (-1) (5)
(b) -5 (Answer)
(11) What is the sum of the three expressions?
(12) (i) 12x is the sum of x and the power of 12
(ii) The index of a is 3 in expression 4a3
(iii) The coefficient of x in the basis of above information?
(c) & (iii) are correct.
(14) 9x2, 8x2, 5y2 are three algebraic expressions then—
(a) What is the sum of numerical coefficients of three expressions?
(d) 22 (Answer)
(b) What is the index of the product of the two expression?
(a) 4 (Answer)
(15) x2+y2+z2, x2+y2z2, -x2+y2-z2 are three algebraic expressions. On the basis of this information, answer the following questions (1) to (4)
(1) Which one will be the sum of difference of the first two expressions with the third expression?
x2+y2+z2 x2-y2+z2 -(- x2+y2-z2)
= x2+y2+z2 x2-y2+z2+ x2-y2-z2
= 3 x2-y2+3z2
(b) 3 x2-y2+3z2 (Answer)
(2) What is the coefficient of y2 in the second expression?
(b) -1 (Answer)
(3) What is the sum of the three expressions?
x2+y2+z2+ x2-y2+z2– x2+y2-z2
(c) x2+y2+z2 (Answer)
(4) Which is the sum of the three expressions?
(16) 3a+4b, a+3b
(17) 2a+ 3b+ 3a+ 5b+ 5a+ 6b
= 10a+ 14b
(18) 4a-3b+(-3a+b) + 2a+3b
(19) 7x+ 5y+ 2z+3x- 6y+7z- 9x+4y +z
= x- 3y+10z
= 6x2+ 6xy- 2z
= 2 (3x2+ 3xy- z) (Answer)
= -2p2+ 15q2 +6r2 (Answer)
(22) 3a +2b- 6c- 5b +4a+ 3c+ 8b -6a +4c
= a+5b +c (Answer)
(23) 2x3-9x2+11x+ 5-x3+7x2 -8x-3-x3+2x2 -4x+1
= -x+3 (Answer)
(24) 5ax+ 3by- 14cz- 11by -7ax- 9cz +3ax+ 6by- 8cz
= a x- 31cz -2by (Answer)
= 5x (Answer)
(26) L.H.S, a2+b2+c2
= x2+y2 -z2+y2+ z2 –x2+ x2-y2
= x+ y+ z
Therefore, L.H.S = R.H.S (Proved)
(27) L.H.S, x+ y+ z
= 5a+7b + 9c+b- 3a-4c +c-2b+a
=3a+ 6b+ 6c
= 3 (a+ 2b+ 2c)
= R.H.S =L.H.S (Proved)
L.H.S, a+ b-c
Therefore = R.H.S (proved)
L.H.S, x-y +z
= a+ b -b- c+ c+ a
Therefore, L.H.S=R.H.S (proved)
L.H.S, x-y+ z
= a+ b +c-a +b +c +b- c +a
= a+ 3b +c
Therefore, L.H.S = R.H.S (proved)
(a) Numerical co-officient of b2 is a 1.
(c) 3a2-2b2+4c2 (Answer)
(40) Solution:- (a) Total price
= (3Xx+2xy) Rs
= (3x+2y) Rs
(b)Solution:- Total price of 5 note book & 8 pencils = (5x+8z) Rs.
Therefore, Algebraic expression [(5x+8z)- 10y]
(c) Solution:- Numeric co-offcient y & z are -2 &5.
Product of numerical co-officient
= 3x (-2) x 5
= -30 (Answer)
(41) (a) Solution:- There are three terms in the first expression.
That’s are—5x2, x y, 3y2
-x2+ 10xy +y2
5x2+ 3xy +4y2
Therefore, Co-officent of x y is 3.
(c) Solution:-(5x2+xy+3y2) – (x2-8xy) – (y2-x2+10xy)
= 5x2+ x y+ 3y2-x2+8xy –y2+x2-10xy
When, x=2: y=1
Therefore, 5x (22) – 2×1 + 2×12
= 20 (Answer)