Class 8: Maths Chapter 6 solutions. Complete Class 8 Maths Chapter 6 Notes.
Contents
RD Sharma Solutions for Class 8 Maths Chapter 6–Algebraic Expressions and Identities
RD Sharma 8th Maths Chapter 6, Class 8 Maths Chapter 6 solutions
EXERCISE 6.1 PAGE NO: 6.2
1. Identify the terms, their coefficients for each of the following expressions:
(i) 7x2yz – 5xy
(ii) x2 + x + 1
(iii) 3x2y2 – 5x2y2z2 + z2
(iv) 9 – ab + bc – ca
(v) a/2 + b/2 – ab
(vi) 0.2x – 0.3xy + 0.5y
Solution:
(i) 7x2yz – 5xy
The given equation has two terms that are:
7x2yz and – 5xy
The coefficient of 7x2yz is 7
The coefficient of – 5xy is – 5
(ii) x2 + x + 1
The given equation has three terms that are:
x2, x, 1
The coefficient of x2 is 1
The coefficient of x is 1
The coefficient of 1 is 1
(iii) 3x2y2 – 5x2y2z2 + z2
The given equation has three terms that are:
3x2y, -5x2y2z2 and z2
The coefficient of 3x2y is 3
The coefficient of -5x2y2z2 is -5
The coefficient of z2 is 1
(iv) 9 – ab + bc – ca
The given equation has four terms that are:
9, -ab, bc, -ca
The coefficient of 9 is 9
The coefficient of -ab is -1
The coefficient of bc is 1
The coefficient of -ca is -1
(v) a/2 + b/2 – ab
The given equation has three terms that are:
a/2, b/2, -ab
The coefficient of a/2 is 1/2
The coefficient of b/2 is 1/2
The coefficient of -ab is -1
(vi) 0.2x – 0.3xy + 0.5y
The given equation has three terms that are:
0.2x, -0.3xy, 0.5y
The coefficient of 0.2x is 0.2
The coefficient of -0.3xy is -0.3
The coefficient of 0.5y is 0.5
2. Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any category?
(i) x+y
(ii) 1000
(iii) x+x2+x3+x4
(iv) 7+a+5b
(v) 2b-3b2
(vi) 2y-3y2+4y3
(vii) 5x-4y+3x
(viii) 4a-15a2
(ix) xy+yz+zt+tx
(x) pqr
(xi) p2q+pq2
(xii) 2p+2q
Solution:
(i) x+y
The given expression contains two terms x and y
∴ It is Binomial
The given expression contains one term 1000
∴ It is Monomial
The given expression contains four terms
∴ It belongs to none of the categories
The given expression contains three terms
∴ It is Trinomial
The given expression contains two terms
∴ It is Binomial
The given expression contains three terms
∴ It is Trinomial
The given expression contains three terms
∴ It is Trinomial
The given expression contains two terms
∴ It is Binomial
The given expression contains four terms
∴ It belongs to none of the categories
The given expression contains one term
∴ It is Monomial
The given expression contains two terms
∴ It is Binomial
The given expression contains two terms
∴ It is Binomial
EXERCISE 6.2 PAGE NO: 6.5
1. Add the following algebraic expressions:
(i) 3a2b, -4a2b, 9a2b
(ii) 2/3a, 3/5a, -6/5a
(iii) 4xy2 – 7x2y, 12x2y -6xy2, -3x2y + 5xy2
(iv) 3/2a – 5/4b + 2/5c, 2/3a – 7/2b + 7/2c, 5/3a + 5/2b – 5/4c
(v) 11/2xy + 12/5y + 13/7x, -11/2y – 12/5x – 137xy
(vi) 7/2x3 – 1/2x2 + 5/3, 3/2x3 + 7/4x2 – x + 1/3, 3/2x2 -5/2x -2
Solution:
(i) 3a2b, -4a2b, 9a2b
Let us add the given expression
3a2b + (-4a2b) + 9a2b
3a2b – 4a2b + 9a2b
8a2b
(ii) 2/3a, 3/5a, -6/5a
Let us add the given expression
2/3a + 3/5a + (-6/5a)
2/3a + 3/5a – 6/5a
Let us take LCM for 3 and 5 which is 15
(2×5)/(3×5)a + (3×3)/(5×3)a – (6×3)/(5×3)a
10/15a + 9/15a – 18/15a
(10a+9a-18a)/15
a/15
(iii) 4xy2 – 7x2y, 12x2y -6xy2, -3x2y + 5xy2
Let us add the given expression
4xy2 – 7x2y + 12x2y – 6xy2 – 3x2y + 5xy2
Upon rearranging
12x2y – 3x2y – 7x2y – 6xy2 + 5xy2 + 4xy2
3xy2 + 2x2y
(iv) 3/2a – 5/4b + 2/5c, 2/3a – 7/2b + 7/2c, 5/3a + 5/2b – 5/4c
Let us add the given expression
3/2a – 5/4b + 2/5c + 2/3a – 7/2b + 7/2c + 5/3a + 5/2b – 5/4c
Upon rearranging
3/2a + 2/3a + 5/3a – 5/4b – 7/2b + 5/2b + 2/5c + 7/2c – 5/4c
By taking LCM for (2 and 3 is 6), (4 and 2 is 4), (5,2 and 4 is 20)
(9a+4a+10a)/6 + (-5b-14b+10b)/4 + (8c+70c-25c)/20
23a/6 – 9b/4 + 53c/20
(v) 11/2xy + 12/5y + 13/7x, -11/2y – 12/5x – 13/7xy
Let us add the given expression
11/2xy + 12/5y + 13/7x + -11/2y – 12/5x – 13/7xy
Upon rearranging
11/2xy – 13/7xy + 13/7x – 12/5x + 12/5y -11/2y
By taking LCM for (2 and 7 is 14), (7 and 5 is 35), (5 and 2 is 10)
(11xy-12xy)/14 + (65x-84x)/35 + (24y-55y)/10
51xy/14 – 19x/35 – 31y/10
(vi) 7/2x3 – 1/2x2 + 5/3, 3/2x3 + 7/4x2 – x + 1/3, 3/2x2 -5/2x – 2
Let us add the given expression
7/2x3 – 1/2x2 + 5/3 + 3/2x3 + 7/4x2 – x + 1/3 + 3/2x2 -5/2x – 2
Upon rearranging
7/2x3 + 3/2x3 – 1/2x2 + 7/4x2 + 3/2x2 – x – 5/2x + 5/3 + 1/3 – 2
10/2x3 + 11/4x2 – 7/2x + 0/6
5x3 + 11/4x2 -7/2x
2. Subtract:
(i) -5xy from 12xy
(ii) 2a2 from -7a2
(iii) 2a-b from 3a-5b
(iv) 2x3 – 4x2 + 3x + 5 from 4x3 + x2 + x + 6
(v) 2/3y3 – 2/7y2 – 5 from 1/3y3 + 5/7y2 + y – 2
(vi) 3/2x – 5/4y – 7/2z from 2/3x + 3/2y – 4/3z
(vii) x2y – 4/5xy2 + 4/3xy from 2/3x2y + 3/2xy2 – 1/3xy
(viii) ab/7 -35/3bc + 6/5ac from 3/5bc – 4/5ac
Solution:
(i) -5xy from 12xy
Let us subtract the given expression
12xy – (- 5xy)
5xy + 12xy
17xy
(ii) 2a2 from -7a2
Let us subtract the given expression
(-7a2) – 2a2
-7a2 – 2a2
-9a2
(iii) 2a-b from 3a-5b
Let us subtract the given expression
(3a – 5b) – (2a – b)
3a – 5b – 2a + b
a – 4b
(iv) 2x3 – 4x2 + 3x + 5 from 4x3 + x2 + x + 6
Let us subtract the given expression
(4x3 + x2 + x + 6) – (2x3 – 4x2 + 3x + 5)
4x3 + x2 + x + 6 – 2x3 + 4x2 – 3x – 5
2x3 + 5x2 – 2x + 1
(v) 2/3y3 – 2/7y2 – 5 from 1/3y3 + 5/7y2 + y – 2
Let us subtract the given expression
1/3y3 + 5/7y2 + y – 2 – 2/3y3 + 2/7y2 + 5
Upon rearranging
1/3y3 – 2/3y3 + 5/7y2 + 2/7y2 + y – 2 + 5
By grouping similar expressions we get,
-1/3y3 + 7/7y2 + y + 3
-1/3y3 + y2 + y + 3
(vi) 3/2x – 5/4y – 7/2z from 2/3x + 3/2y – 4/3z
Let us subtract the given expression
2/3x + 3/2y – 4/3z – (3/2x – 5/4y – 7/2z)
Upon rearranging
2/3x – 3/2x + 3/2y + 5/4y – 4/3z + 7/2z
By grouping similar expressions we get,
LCM for (3 and 2 is 6), (2 and 4 is 4), (3 and 2 is 6)
(4x-9x)/6 + (6y+5y)/4 + (-8z+21z)/6
-5x/6 + 11y/4 + 13z/6
(vii) x2y – 4/5xy2 + 4/3xy from 2/3x2y + 3/2xy2 – 1/3xy
Let us subtract the given expression
2/3x2y + 3/2xy2 – 1/3xy – (x2y – 4/5xy2 + 4/3xy)
Upon rearranging
2/3x2y – x2y + 3/2xy2 + 4/5xy2 – 1/3xy – 4/3xy
By grouping similar expressions we get,
LCM for (3 and 1 is 3), (2 and 5 is 10), (3 and 3 is 3)
-1/3x2y + 23/10xy2 – 5/3xy
(viii) ab/7 -35/3bc + 6/5ac from 3/5bc – 4/5ac
Let us subtract the given expression
3/5bc – 4/5ac – (ab/7 – 35/3bc + 6/5ac)
Upon rearranging
3/5bc + 35/3bc – 4/5ac – 6/5ac – ab/7
By grouping similar expressions we get,
LCM for (5 and 3 is 15), (5 and 5 is 5)
(9bc+175bc)/15 + (-4ac-6ac)/5 – ab/7
184bc/15 + -10ac/5 – ab/7
– ab/7 + 184bc/15 – 2ac
3. Take away:
(i) 6/5x2 – 4/5x3 + 5/6 + 3/2x from x3/3 – 5/2x2 + 3/5x + 1/4
(ii) 5a2/2 + 3a3/2 + a/3 – 6/5 from 1/3a3 – 3/4a2 – 5/2
(iii) 7/4x3 + 3/5x2 + 1/2x + 9/2 from 7/2 – x/3 – x2/5
(iv) y3/3 + 7/3y2 + 1/2y + 1/2 from 1/3 – 5/3y2
(v) 2/3ac – 5/7ab + 2/3bc from 3/2ab -7/4ac – 5/6bc
Solution:
(i) 6/5x2 – 4/5x3 + 5/6 + 3/2x from x3/3 – 5/2x2 + 3/5x + 1/4
Let us subtract the given expression
1/3x3 – 5/2x2 + 3/5x + 1/4 – (6/5x2 – 4/5x3 + 5/6 + 3/2x)
Upon rearranging
1/3x3 + 4/5x3 – 5/2x2 – 6/5x2 + 3/5x – 3/2x + 1/4 – 5/6
By grouping similar expressions we get,
LCM for (3 and 5 is 15), (2 and 5 is 10), (5 and 2 is 10), (4 and 6 is 24)
17/15x3 – 37/10x2 – 9/10x – 14/24
17/15x3 – 37/10x2 – 9/10x – 7/12
(ii) 5a2/2 + 3a3/2 + a/3 – 6/5 from 1/3a3 – 3/4a2 – 5/2
Let us subtract the given expression
1/3a3 – 3/4a2 – 5/2 – (5/2a2 + 3/2a3 + a/3 – 6/5)
Upon rearranging
1/3a5 – 3/2a3 – 3/4a2 – 5/2a2 – a/3 – 5/2 + 6/5
By grouping similar expressions we get,
LCM for (3 and 2 is 6), (4 and 2 is 4), (2 and 5 is 10)
(2a3 – 9a3)/6 – (3a2 + 10a2)/4 – a/3 + (-25+12)/10
-7/6a3 – 13/4a2 – a/3 – 13/10
(iii) 7/4x3 + 3/5x2 + 1/2x + 9/2 from 7/2 – x/3 – x2/5
Let us subtract the given expression
7/2 – x/3 – 1/5x2 – (7/4x3 + 3/5x2 + 1/2x + 9/2)
Upon rearranging
-7/4x3 – 1/5x2 – 3/5x2 – x/3 – x/2 + 7/2 – 9/2
By grouping similar expressions we get,
LCM for (3 and 2 is 6)
-7/4x3 – 4/5x2– (2x-3x)/6 + (7-9)/2
-7/4x3 – 4/5x2 – 5/6x – 1
(iv) y3/3 + 7/3y2 + 1/2y + 1/2 from 1/3 – 5/3y2
Let us subtract the given expression
1/3 – 5/3y2 – (1/3y3 + 7/3y2 + 1/2y + 1/2)
Upon rearranging
-1/3y3 – 5/3y2 – 7/3y2 – 1/2y + 1/3 – 1/2
By grouping similar expressions we get,
LCM for (3 and 3 is 3), (3 and 2 is 6)
-1/3y3 + (-5y2 – 7y2)/3 – 1/2y + (2-3)/6
-1/3y3 – 12/3y2 – 1/2y – 1/6
(v) 2/3ac – 5/7ab + 2/3bc from 3/2ab -7/4ac – 5/6bc
Let us subtract the given expression
3/2ab – 7/4ac – 5/6bc – (2/3ac – 5/7ab + 2/3bc)
Upon rearranging
3/2ab + 5/7ab – 7/4ac – 2/3ac – 5/6bc – 2/3bc
By grouping similar expressions we get,
LCM for (2 and 7 is 14), (4 and 3 is 12), (6 and 3 is 6)
(21ab+10ab)/14 – (21ac-8ac)/12 – (5bc-4bc)/6
31/14ab – 29/12ac – 3/2bc
4. Subtract 3x – 4y – 7z from the sum of x – 3y + 2z and -4x + 9y – 11z.
Solution:
The sum of x – 3y + 2z and -4x + 9y – 11z is
(x – 3y + 2z) + (-4x + 9y – 11z)
Upon rearranging
x – 4x – 3y + 9y + 2z – 11z
-3x + 6y – 9z
Now, Let us subtract the given expression from -3x + 6y – 9z
(-3x + 6y – 9z) – (3x – 4y – 7z)
Upon rearranging
-3x – 3x + 6y + 4y – 9z + 7z
-6x + 10y – 2z
5. Subtract the sum of 3l – 4m – 7n2 and 2l + 3m – 4n2 from the sum of 9l + 2m – 3n2 and -3l + m + 4n2….
Solution:
Sum of 3l – 4m – 7n2 and 2l + 5m – 4n2
3l – 4m – 7n2 + 2l + 3m – 4n2
Upon rearranging
3l + 2l – 4m + 3m – 7n2 – 4n2
5l – m – 11n2 ……………………..equation (1)
Sum of 9l + 2m – 3n2 and -3l + m + 4n2
9l + 2m – 3n2 + (-3l + m + 4n2)
Upon rearranging
9l – 3l + 2m + m – 3n2 + 4n2
6l + 3m + n2 ……………………….equation (2)
Let us subtract equation (i) from (ii), we get
6l + 3m + n2 – (5l – m – 11n2)
Upon rearranging
6l – 5l + 3m + m + n2 + 11n2
l + 4m + 12n2
6. Subtract the sum of 2x – x2 + 5 and -4x – 3 + 7x2 from 5.
Solution:
Sum of 2x – x2 + 5 and -4x – 3 + 7x2 is
2x – x2 + 5 + (-4x – 3 + 7x2)
2x – x2 + 5 – 4x – 3 + 7x2
Upon rearranging
– x2 + 7x2 + 2x – 4x + 5 – 3
6x2 -2x + 2 ………….equation (i)
Let us subtract equation (i) from 5 we get,
5 – (6x2 -2x + 2)
5 – 6x2 + 2x – 2
3 + 2x – 6x2
7. Simplify each of the following:
(i) x2 – 3x + 5 – 1/2(3x2 – 5x + 7)
(ii) [5 – 3x + 2y – (2x – y)] – (3x – 7y + 9)
(iii) 11/2x2y – 9/4xy2 + 1/4xy – 1/14y2x + 1/15yx2 + 1/2xy
(iv) (1/3y2 – 4/7y + 11) – (1/7y – 3 + 2y2) – (2/7y – 2/3y2 + 2)
(v) -1/2a2b2c + 1/3ab2c – 1/4abc2 – 1/5cb2a2 + 1/6cb2a – 1/7c2ab + 1/8ca2b
Solution:
(i) x2 – 3x + 5 – 1/2(3x2 – 5x + 7)
Upon rearranging
x2 – 3/2x2 – 3x + 5/2x + 5 – 7/2
By grouping similar expressions we get,
LCM for (1 and 2 is 2)
(2x2 – 3x2)/2 – (6x + 5x)/2 + (10-7)/2
-1/2x2 – 1/2x + 3/2
(ii) [5 – 3x + 2y – (2x – y)] – (3x – 7y + 9)
5 – 3x + 2y – 2x + y – 3x + 7y – 9
Upon rearranging
– 3x – 2x – 3x + 2y + y + 7y + 5 – 9
By grouping similar expressions we get,
-8x + 10y – 4
(iii) 11/2x2y – 9/4xy2 + 1/4xy – 1/14y2x + 1/15yx2 + 1/2xy
Upon rearranging
11/2x2y + 1/15x2y – 9/4xy2 – 1/14xy2 + 1/4xy + 1/2xy
By grouping similar expressions we get,
LCM for (2 and 15 is 30), (4 and 14 is 56), (4 and 2 is 4)
(165x2y + 2x2y)/30 + (-126xy2 – 4xy2)/56 + (xy + 2xy)/4
167/30x2y – 130/56xy2 + 3/4xy
167/30x2y – 65/28xy2 + 3/4xy
(iv) (1/3y2 – 4/7y + 11) – (1/7y – 3 + 2y2) – (2/7y – 2/3y2 + 2)
Upon rearranging
1/3y2 – 2y2 – 2/3y2 – 4/7y – 1/7y – 2/7y + 11 + 3 – 2
By grouping similar expressions we get,
LCM for (3, 1 and 3 is 3), (7, 7 and 7 is 7)
(y2 – 6y2 + 2y2)/3 – (4y – y – 2y)/7 + 12
-3/3y2 – 7/7y + 12
-y2 – y + 12
(v) -1/2a2b2c + 1/3ab2c – 1/4abc2 – 1/5cb2a2 + 1/6cb2a – 1/7c2ab + 1/8ca2b
Upon rearranging
-1/2a2b2c – 1/5a2b2c + 1/3ab2c + 1/6ab2c – 1/4abc2 – 1/7abc2 + 1/8a2bc
By grouping similar expressions we get,
LCM for (2 and 5 is 10), (3 and 6 is 6), (4 and 7 is 28)
-7/10a2b2c + 1/2ab2c – 11/28abc2 + 1/8a2bc
EXERCISE 6.3 PAGE NO: 6.13
Find each of the following products:
1. 5x2 × 4x3
Solution:
Let us simplify the given expression
5 × x × x × 4 × x × x × x
5 × 4 × x1+1+1+1+1
20 × x5
20x5
2. -3a2 × 4b4
Solution:
Let us simplify the given expression
– 3 × a2 × 4 × b4
-12 × a2 × b4
-12a2b4
3. (-5xy) × (-3x2yz)
Solution:
Let us simplify the given expression
(-5) × (-3) × x × x2 × y × y × z
15 × x1+2 × y1+1 × z
15x3y2z
4. 1/2xy × 2/3x2yz2
Solution:
Let us simplify the given expression
1/2 × 2/3 × x × x2 × y × y × z2
1/3 × x1+2 × y1+1 × z2
1/3x3y2z2
5. (-7/5xy2z) × (13/3x2yz2)
Solution:
Let us simplify the given expression
-7/5 × 13/3 × x × x2 × y2 × y × z × z2
-91/15 × x1+2 × y2+1 × z1+2
-91/15x3y3z3
6. (-24/25x3z) × (-15/16xz2y)
Solution:
Let us simplify the given expression
-24/25 × -15/16 × x3 × x × z × z2 × y
18/20 × x3+1 × z1+2 × y
9/10x4z3y
7. (-1/27a2b2) × (9/2a3b2c2)
Solution:
Let us simplify the given expression
-1/27 × 9/2 × a2 × a3 × b2 × b2 × c2
-1/6 x a2+3 × b2+2 × c2
-1/6a5b4c2
8. (-7xy) × (1/4x2yz)
Solution:
Let us simplify the given expression
-7 × 1/4 × x × y × x2 × y × z
-7/4 × x1+2 × y1+1 × z
-7/4x3y2z
9. (7ab) × (-5ab2c) × (6abc2)
Solution:
Let us simplify the given expression
7 × -5 × 6 × a × a × a × b × b2 × b × c × c2
210 × a1+1+1 × b1+2+1 × c1+2
210a3b4c3
10. (-5a) × (-10a2) × (-2a3)
Solution:
Let us simplify the given expression
(-5) × (-10) × (-2) × a × a2 × a3
-100 × a1+2+3
-100a6
11. (-4x2) × (-6xy2) × (-3yz2)
Solution:
Let us simplify the given expression
(-4) × (-6) – (-3) × x2 × x × y2 × y × z2
– 72 × x2+1 × y2+1 × z2
-72x3y3z2
12. (-2/7a4) × (-3/4a2b) × (-14/5b2)
Solution:
Let us simplify the given expression
-2/7 × -3/4 × -14/5 × a4 × a2 × b × b2
-6/10 × a4+2 × b1+2
-3/5a6b3
13. (7/9ab2) × (15/7ac2b) × (-3/5a2c)
Solution:
Let us simplify the given expression
7/9 × 15/7 × -3/5 × a × a × a2 × b2 × b × c2 × c
– a1+1+2 × b2+1 × c2+1
-a4b3c3
14. (4/3u2vw) × (-5uvw2) × (1/3v2wu)
Solution:
Let us simplify the given expression
4/3 × -5 × 1/3 × u2 × u × u × v × v × v2 × w × w2 × w
-20/9 × u2+1+1 × v1+1+2 × w1+2+1
-20/9u4v4w4
15. (0.5x) × (1/3xy2z4) × (24x2yz)
Solution:
Let us simplify the given expression
0.5 × 1/3 × 24 × x × x × y2 × y × x2 × z4 × z
12/3 × x1+1+2 × y2+1 × z4+1
4x4 × y3 × z5
4x4y3z5
16. (4/3pq2) × (-1/4p2r) × (16p2q2r2)
Solution:
Let us simplify the given expression
4/3 × 1/4 × 16 × p × p2 × p2 × q2 × q2 × r × r2
-16/3 × p1+2+2 × q2+2 × r1+2
-16/3p5q4r3
17. (2.3xy) × (0.1x) × (0.16)
Solution:
Let us simplify the given expression
2.3 × 0.1 × 0.16 × x × x × y
0.0368 × x1+1 × y
0.0368x2y
Express each of the following products as a monomials and verify the result in each case for x=1:
18. (3x) × (4x) × (-5x)
Solution:
Let us simplify the given expression
3 × 4 × -5 × x × x × x
-60 × x1+1+1
-60x3
Verification
LHS = (3 × 1) × (4 × 1) × (-5 × 1)
= 3 × 4 × – 5
= – 60
RHS = -60 (1)3 = – 60
Therefore, LHS = RHS.
19. (4x2) × (-3x) × (4/5x3)
Solution:
Let us simplify the given expression
4 × -3 × 4/5 × x2 × x × x3
-48/5 × x2+1+3
-48/5x6
Verification
LHS = 4 × 12 × – 3 × 1 × 4/5 × 13
= – 48/5
RHS = – 48/5 × 16 = – 48/5
Therefore, LHS = RHS.
20. (5x4) × (x2)3 × (2x) 2
Solution:
Let us simplify the given expression
5 × x4 × x6 × 4 × x2
5 × 4 × x4 × x6 × x2
20 × x4+6+2
20x12
Verification
LHS = (5 × 14) × (12)3 × (2 × 1)2
= 5 × 4
= 20
RHS = 20 × 112 = 20
Therefore, LHS = RHS.
21. (x2)3 × (2x) × (-4x) × (5)
Solution:
Let us simplify the given expression
x6 × 2 × x × -4 × x × 5
2 × -4 × 5 × x6 × x × x
-40 × x6+1+1
-40x8
Verification
LHS = (12)3 × (2 × 1) × (-4 × 1) × 5
= – 40
RHS = – 40 × 18 = – 40
Therefore, LHS = RHS.
22. Write down the product of -8x2y6 and -20xy verify the product for x = 2.5, y = 1
Solution:
Let us simplify the given expression
-8 × -20 × x2 × x × y6 × y
160 × x2+1 × y6+1
160x3y7
Now let us verify when, x = 2.5 and y = 1
For 160x3y7
160 (2.5)3 × (1)7
160 × 15.625
2500
For -8x2y6 and -20xy
-8 × 2.52 × 16 × -20 × 1 × 2.5
2500
Hence, the given expression is verified.
23. Evaluate (3.2x6y3) × (2.1x2y2) when x = 1 and y = 0.5
Solution:
Let us simplify the given expression
3.2 × 2.1 × x6 × x2 × y3 × y2
6.72 × x6+2 × y3+2
6.72x8y5
Now let us substitute when, x = 1 and y = 0.5
For 6.72x8y5
6.72 × 18 × 0.55
0.21
24. Find the value of (5x6) × (-1.5x2y3) × (-12xy2) when x = 1, y = 0.5
Solution:
Let us simplify the given expression
5 × -1.5 × -12 × x6 × x2 × x × y3 × y2
90 × x6+2+1 × y3+2
90x9y5
Now let us substitute when, x = 1 and y = 0.5
For 90x9y5
90 × (1)9× (0.5)5
2.8125
45/16
25. Evaluate (2.3a5b2) × (1.2a2b2) when a = 1 and b = 0.5
Solution:
Let us simplify the given expression
2.3a5b2 × 1.2a2b2
2.3 × 1.2 × a5 × a2 × b2 × b2
2.76 × a5+2 × b2+2
2.76a7b4
Now let us substitute when, a = 1 and b = 0.5
For 2.76 a7 b4
2.76 (1)7 (0.5)4
2.76 × 1 × 0.0025
0.1725
6.9/40
26. Evaluate (-8x2y6) × (-20xy) for x = 2.5 and y = 1
Solution:
Let us simplify the given expression
-8 × – 20 × x2 × x × y6 × y
160x2+1y6+1
160x3y7
Now let us substitute when, x = 2.5 and y = 1
160x3y7
160 × (2.5)3 × (1)7
2500
Express each of the following products as a monomials and verify the result for x = 1, y = 2:
27. (-xy3) × (yx3) × (xy)
Solution:
Let us simplify the given expression
-x × y3 × y × x3 × x × y
-x1+3+1 × y3+1+1
-x5y5
Now let us substitute when, x = 1 and y = 2
-x5y5
-15 × 25
-32
28. (1/8x2y4) × (1/4x4y2) × (xy) × 5
Solution:
Let us simplify the given expression
1/8 × 1/4 × 5 × x2 × x4 × x × y4 × y2 × y
5/32 × x2+4+1 × y4+2+1
5/32x7y7
Now let us substitute when, x = 1 and y = 2
5/32 × 17 × 27
5/32 × 128
5 × 4
20
29. (2/5a2b) × (-15b2ac) × (-1/2c2)
Solution:
Let us simplify the given expression
2/5 × -15 × -1/2 × a2 × a × b × b2 × c × c2
3 × a2+1 × b1+2 × c1+2
3a3b3c3
30. (-4/7a2b) × (-2/3b2c) × (-7/6c2a)
Solution:
Let us simplify the given expression
-4/7 × -2/3 × -7/6 × a2 × a × b × b2 × c × c2
-4/9 × a2+1 × b2+1 × c1+2
-4/9a3b3c3
31. (4/9abc3) × (-27/5a3b2) × (-8b3c)
Solution:
Let us simplify the given expression
4/9 × -27/5 × -8 × a × a3 × b × b2 × b3 × c3 × c
96/5 × a1+3 × b1+2+3 × c3+1
96/5a4b6c4
Evaluate each of the following when x = 2, y = -1.
32. (2xy) × (x2y/4) × (x2) × (y2)
Solution:
Let us simplify the given expression
2 × 1/4 × x × x2 × x2 × y × y2 × y
1/2x1+2+2y1+2+1
1/2x5y4
Now let us substitute when, x = 2 and y = -1
For 1/2x5y4
1/2 × (2)5 × (-1)4
1/2 × 32 × 1
16
33. (3/5x2y) × (-15/4xy2) × (7/9x2y2)
Solution:
Let us simplify the given expression
3/5 × -15/4 × 7/9 × x2 × x × x2 × y × y2 × y2
-7/4 × x2+1+2 × y1+2+2
7/4x5y5
Now let us substitute when, x = 2 and y = -1
For -7/4x5y5
-7/4 × (2)5 (-1)5
-7/4 × 32 × -1
56
EXERCISE 6.4 PAGE NO: 6.21
Find the following products:
1. 2a3 (3a + 5b)
Solution:
Let us simplify the given expression
2a3 (3a + 5b)
(2a3 × 3a) + (2a3 × 5b)
6a3+1 + 10a3b
6a4 + 10a3b
2. -11a (3a + 2b)
Solution:
Let us simplify the given expression
-11a (3a + 2b)
(-11a × 3a) + (-11a × 2b)
-33a2 – 22ab
3. -5a (7a – 2b)
Solution:
Let us simplify the given expression
-5a (7a – 2b)
(-5a × 7a) – (-5a × 2b)
-35a2 + 10ab
4. -11y2 (3y + 7)
Solution:
Let us simplify the given expression
-11y2 (3y + 7)
(-11y2 × 3y) + (-11y2 × 7)
-33y3 – 77y2
5. 6x/5(x3 + y3)
Solution:
Let us simplify the given expression
6/5x (x3 + y3)
(6/5x × x3) + (6/5x × y3)
6/5x4 + 6/5xy3
6. xy (x3 – y3)
Solution:
Let us simplify the given expression
xy (x3 – y3)
(xy × x3) – (xy × y3)
x4y – xy4
7. 0.1y (0.1x5 + 0.1y)
Solution:
Let us simplify the given expression
0.1y (0.1x5 + 0.1y)
(0.1y × 0.1x5) + (0.1y × 0.1y)
0.01x5y + 0.01y2
8. (-7/4ab2c – 6/25a2c2) (-50a2b2c2)
Solution:
Let us simplify the given expression
(-7/4ab2c – 6/25a2c2) (-50a2b2c2)
(-7/4ab2c × -50a2b2c2) – (6/25a2c2 × -50a2b2 × c2)
350/4a3b4c3 + 12a4b2c4
175/2a3b4c3 + 12a4b2c4
9. -8/27xyz (3/2xyz2 – 9/4xy2z3)
Solution:
Let us simplify the given expression
-8/27xyz (3/2xyz2 – 9/4xy2z3)
(-8/27xyz × 3/2xyz2) – (-8/27xyz × 9/4xy2z3)
-4/9x2y2z3 + 2/3x2y3z4
10. -4/27xyz (9/2x2yz – 3/4xyz2)
Solution:
Let us simplify the given expression
-4/27xyz (9/2x2yz – 3/4xyz2)
(-4/27xyz × 9/2x2yz) – (-4/27xyz × 3/4xyz2)
-2/3x3y2z2 + 1/9x2y2z3
11. 1.5x (10x2y – 100xy2)
Solution:
Let us simplify the given expression
1.5x (10x2y – 100xy2)
(1.5x × 10x2y) – (1.5x × 100xy2)
15x3y – 150x2y2
12. 4.1xy (1.1x – y)
Solution:
Let us simplify the given expression
4.1xy (1.1x – y)
(4.1xy × 1.1x) – (4.1xy × y)
4.51x2y – 4.1xy2
13. 250.5xy (xz + y/10)
Solution:
Let us simplify the given expression
250.5xy (xz + y/10)
(250.5xy × xz) + (250.5xy × y/10)
250.5x2yz + 25.05xy2
14. 7/5x2y (3/5xy2 + 2/5x)
Solution:
Let us simplify the given expression
7/5x2y (3/5xy2 + 2/5x)
(7/5x2y × 3/5xy2) + (7/5x2y × 2/5x)
21/25x3y3 + 14/25x3y
15. 4/3a (a2 + b2 – 3c2)
Solution:
Let us simplify the given expression
4/3a (a2 + b2 – 3c2)
(4/3a × a2) + (4/3a × b2) – (4/3a × 3c2)
4/3a3 + 4/3ab2 – 4ac2
16. Find the product 24x2 (1-2x) and evaluate its value for x = 3
Solution:
Let us simplify the given expression
24x2 (1 – 2x)
(24x2× 1) – (24x2× 2x)
24x2 – 48x3
Now let us evaluate the expression when x = 3
24x2 – 48x3
24 × (3)2 – 48 × (3)3
24 × (9) – 48 × (27)
216 – 1296
-1080
17. Find the product -3y (xy+y2) and evaluate its value for x = 4 and y = 5
Solution:
Let us simplify the given expression
-3y (xy+y2)
(-3y × xy) + (-3y × y2)
-3xy2 – 3y3
Now let us evaluate the expression when x = 4 and y = 5
-3xy2 – 3y3
-3 × (4) × (5)2 – 3 × (5)3
-300 – 375
-675
18. Multiply -3/2x2y3 by (2x-y) and verify the answer for x = 1 and y = 2
Solution:
Let us simplify the given expression
-3/2x2y3 by (2x-y)
(-3/2x2y3 × 2x) – (-3/2x2y3 × y)
-3x3y3 + 3/2x2y4
Now let us evaluate the expression when x = 1 and y = 2
-3x3y3 + 3/2x2y4
-3 × (1)4 × (2)3 + 3/2 × (1)2 × (2)4
– 3 × (8) + 3 (8)
-24+24
0
19. Multiply the monomial by the binomial and find the value of each for x = -1, y = 0.25 and z = 0.005:
(i) 15y2 (2 – 3x)
(ii) -3x (y2 + z2)
(iii) z2 (x – y)
(iv) xz (x2 + y2)
Solution:
(i) 15y2 (2 – 3x)
Let us simplify the given expression
30y2 – 45xy2
By evaluating the values in the expression x = -1, y = 25/100 and z = 5/1000
30 × (25/100)2 – 45 × (-1) × (25/100)2
30 (1/16) + 45 (1/16)
15/8 + 45/16
(30+45)/16
75/16
(ii) -3x (y2 + z2)
Let us simplify the given expression
-3xy2 + -3xz2
By evaluating the values in the expression x = -1, y = 25/100 and z = 5/1000
-3× (-1) × (25/100)2 – 3 × (-1) × (5/1000)2
(3×25×25/100×100) + (3×5×5/1000×1000)
3/16 + 3/40000
39/200
(iii) z2 (x – y)
Let us simplify the given expression
z2x – z2y
By evaluating the values in the expression x = -1, y = 25/100 and z = 5/1000
z2 (x – y)
(5/1000)2 (-1 – 25/100)
(1/40000) (-100-25/100)
(1/40000) (-125/100)
(1/40000) (-5/4)
-5/160000
-1/32000
(iv) xz (x2 + y2)
Let us simplify the given expression
x3z + xzy2
By evaluating the values in the expression x = -1, y = 25/100 and z = 5/1000
x3z + xzy2
(-1)3 × (5/1000) + (-1) × (5/1000) × (25/100)2
-1/200 – 1/16 × 1/200
-1/200 – 1/3200
By taking LCM as 3200
(-16 -1)/3200
-17/3200
20. Simplify:
(i) 2x2 (x3 – x) – 3x (x4 + 2x) – 2 (x4 – 3x2)
(ii) x3y (x2 – 2x) + 2xy (x3 – x4)
(iii) 3a2 + 2 (a+2) – 3a (2a+1)
(iv) x (x+4) + 3x (2x2 -1) + 4x2 + 4
(v) a (b-c) – b (c-a) – c (a-b)
(vi) a (b-c) +b (c-a) + c (a-b)
(vii) 4ab (a-b) – 6a2 (b-b2) -3b2 (2a2 -a) + 2ab (b-a)
(viii) x2 (x2 + 1) – x3 (x + 1) – x (x3 – x)
(ix) 2a2 + 3a (1 – 2a3) + a (a + 1)
(x) a2 (2a – 1) + 3a + a3 – 8
(xi) 3/2x2 (x2 – 1) + 1/4x2 (x2 + x) – 3/4x (x3 – 1)
(xii) a2b (a-b2) + ab2(4ab – 2a2) – a3b(1-2b)
(xiii) a2b (a3– a + 1) – ab(a4 – 2a2 + 2a) – b(a3– a2 – 1)
Solution:
(i) 2x2 (x3 – x) – 3x (x4 + 2x) – 2 (x4 – 3x2)
Let us simplify the given expression
2x5 – 2x3 – 3x5 – 6x2 – 2x4 + 6x2
By grouping similar expressions we get,
2x5 – 3x5 – 2x3 – 2x4 – 6x2 + 6x2
-x5 – 2x4 – 2x3
(ii) x3y (x2 – 2x) + 2xy (x3 – x4)
Let us simplify the given expression
x5y – 2x4y + 2x4y – 2x5y
By grouping similar expressions we get,
-x5y – 2x5y
-x5y
(iii) 3a2 + 2 (a+2) – 3a (2a+1)
Let us simplify the given expression
3a2 + 2a + 4 – 6a2 – 3a
By grouping similar expressions we get,
3a2 – 6a2 + 2a – 3a + 4
-3a2 – a + 4
(iv) x (x+4) + 3x (2x2 -1) + 4x2 + 4
Let us simplify the given expression
x2 + 4x + 6x3 – 3x + 4x2 + 4
By grouping similar expressions we get,
6x3 + 5x2 + x + 4
(v) a (b-c) – b (c-a) – c (a-b)
Let us simplify the given expression
ab – ac – bc + ab – ca + bc
By grouping similar expressions we get,
2ab – 2ac
(vi) a (b-c) +b (c-a) + c (a-b)
Let us simplify the given expression
ab – ac + bc – ab + ac – bc
By grouping similar expressions we get,
0
(vii) 4ab (a-b) – 6a2 (b-b2) -3b2 (2a2 -a) + 2ab (b-a)
Let us simplify the given expression
4a2b – 4ab2 – 6a2b + 6a2b2 – 6a2b2 + 3ab2 + 2ab2 – 2a2b
By grouping similar expressions we get,
4a2b – 6a2b– 2a2b – 4ab2 + 3ab2 + 2ab2 + 6a2b2 – 6a2b2
-4a2b + ab2
(viii) x2 (x2 + 1) – x3 (x + 1) – x (x3 – x)
Let us simplify the given expression
x4 + x2 – x4 – x3 – x4 + x2
By grouping similar expressions we get,
x4 – x4 – x4 – x3 + x2 + x2
– x4 – x3 + 2x2
(ix) 2a2 + 3a (1 – 2a3) + a (a + 1)
Let us simplify the given expression
2a2 + 3a – 6a4 + a2 + a
By grouping similar expressions we get,
-6a4 + 3a2 + 4a
(x) a2 (2a – 1) + 3a + a3 – 8
Let us simplify the given expression
2a3 – a2 + 3a + a3 – 8
By grouping similar expressions we get,
3a3 – a2 + 3a – 8
(xi) 3/2x2 (x2 – 1) + 1/4x2 (x2 + x) – 3/4x (x3 – 1)
Let us simplify the given expression
3/2x4 – 3/2x2 + 1/4x4 + 1/4x3 – 3/4x4 + 3/4x
By grouping similar expressions we get,
3/2x4 + 1/4x4 – 3/4x4 – 3/2x2 + 1/4x3 + 3/4x
4/4x4 + 1/4x3 – 3/2x2 + 3/4x
x4 + 1/4x3 – 3/2x2 + 3/4x
(xii) a2b (a-b2) + ab2(4ab – 2a2) – a3b(1-2b)
Let us simplify the given expression
a3b – a2b3 + 4a2b3 – 2a3b2 – a3b + 2a3b2
By grouping similar expressions we get,
-a2b3 + 4a2b3
3a2b3
(xiii) a2b (a3– a + 1) – ab(a4 – 2a2 + 2a) – b(a3– a2 – 1)
Let us simplify the given expression
a5b – a3b + a2b – a5b + 2a3b – 2a2b – ba3 + a2b + b
By grouping similar expressions we get,
a5b – a5b – a3b + 2a3b – ba3 + a2b – 2a2b + a2b + b
b
EXERCISE 6.5 PAGE NO: 6.30
Multiply:
1. (5x + 3) by (7x + 2)
Solution:
Now let us simplify the given expression
(5x + 3) × (7x + 2)
5x (7x + 2) + 3 (7x + 2)
35x2 + 10x + 21x + 6
35x2 + 31x + 6
2. (2x + 8) by (x – 3)
Solution:
Now let us simplify the given expression
(2x + 8) × (x – 3)
2x (x – 3) + 8 (x – 3)
2x2 – 6x + 8x – 24
2x2 + 2x – 24
3. (7x + y) by (x + 5y)
Solution:
Now let us simplify the given expression
(7x + y) × (x + 5y)
7x (x + 5y) + y (x + 5y)
7x2 + 35xy + xy + 5y2
7x2 + 36xy + 5y2
4. (a – 1) by (0.1a2 + 3)
Solution:
Now let us simplify the given expression
(a – 1) × (0.1a2 + 3)
a (0.1a2 + 3) -1 (0.1a2 + 3)
0.1a3 + 3a – 0.1a2 – 3
0.1a3 – 0.1a2 + 3a – 3
5. (3x2 + y2) by (2x2 + 3y2)
Solution:
Now let us simplify the given expression
(3x2 + y2) × (2x2 + 3 y2)
3x2 × (2x2 + 3y2) + y2 × (2x2 + 3y2)
6x4 + 9x2y2 + 2x2y2 + 3y4
6x4 + 11x2y2 + 3y4
6. (3/5x + 1/2y) by (5/6x + 4y)
Solution:
Now let us simplify the given expression
(3/5x + 1/2y) × (5/6x + 4y)
3/5x × (5/6x + 4y) + 1/2y × (5/6x + 4y)
15/30x2 + 12/5xy + 5/12xy + 4/2y2
1/2x2 + 169/60xy + 2y2
7. (x6 – y6) by (x2 + y2)
Solution:
Now let us simplify the given expression
(x6 – y6) × (x2 + y2)
x6 × (x2 + y2) – y6 × (x2 + y2)
x8 + x6y2 – x2y6 – y8
8. (x2 + y2) by (3a + 2b)
Solution:
Now let us simplify the given expression
(x2 + y2) × (3a + 2b)
x2 × (3a + 2b) + y2 × (3a + 2b)
3ax2 + 3ay2 + 2bx2 + 2by2
9. (- 3d – 7f) by (5d + f)
Solution:
Now let us simplify the given expression
(- 3d – 7f) × (5d + f)
-3d (5d + f) – 7f (5d + f)
– 15d2 – 3df – 35df – 7f2
– 15d2 – 38df – 7f2
10. (0.8a – o.5b) by (1.5a – 3b)
Solution:
Now let us simplify the given expression
(0.8a – 0.5b) × (1.5a – 3b)
0.8a (1.5a – 3b) – 0.5b (1.5a – 3b)
1.2a2 – 2.4ab – 0.75ab + 1.5b2
1.2a2 – 3.15ab + 1.5b2
11. (2x2y2 – 5xy2) by (x2 – y2)
Solution:
Now let us simplify the given expression
(2x2y2 – 5xy2) × (x2 – y2)
2x2y2 (x2 – y2) – 5xy2 (x2 – y2)
2x4y2 – 5x3y2 – 2x2y4 + 5xy4
12. (x/7 + x2/2) by (2/5 + 9x/4)
Solution:
Now let us simplify the given expression
(x/7 + x2/2) × (2/5 + 9x/4)
x/7 (2/5 + 9x/4) + x2/2 (2/5 + 9x/4)
2x/35 + (9 x2)/28 + x2/5 + (9 x3)/8
9/8x3 + 73/140x2 + 2/35x
13. (-a/7 + a2/9) by (b/2 – b2/3)
Solution:
Now let us simplify the given expression
(-a/7 + a2/9) × (b/2 – b2/3)
-a/7 (b/2 – b2/3) + a2/9 (b/2 – b2/3)
-ab/14 + ab2/21 + a2b/18 – a2b2/27
14. (3x2y – 5xy2) by (1/5x2 + 1/3y2)
Solution:
Now let us simplify the given expression
(3x2y – 5xy2) × (1/5x2 + 1/3y2)
3x2y (1/5x2 + 1/3y2) – 5xy2 (1/5x2 + 1/3y2)
3/5x4y + 3/3x2y3 – x3y2 + 5/3xy4
3/5x4y + x2y3 – x3y2 + 5/3xy4
15. (2x2 – 1) by (4x3 + 5x2)
Solution:
Now let us simplify the given expression
(2x2 – 1) × (4x3 + 5x2)
2x2 (4x3 + 5x2) – 1 (4x3 + 5x2)
8x5 + 10x4 – 4x3 – 5x2
16. (2xy + 3y2) by (3y2 – 2)
Solution:
Now let us simplify the given expression
(2xy + 3y2) × (3y2 – 2)
2xy (3y2 – 2) + 3y2 (3y2 – 2)
6xy3 – 4xy + 9y4 – 6y2
Find the following products and verify the results for x = -1, y = -2:
17. (3x – 5y) (x + y)
Solution:
Now let us simplify the given expression
(3x – 5y) × (x + y)
(3x – 5y) × (x + y)
x (3x – 5y) + y (3x – 5y)
3x2 – 5xy + 3xy – 5y2
3x2 – 2xy – 5y2
Let us substitute the given values x = – 1 and y = – 2, then
(3x – 5y) × (x + y)[3 (-1) – 5 (-2)] × [(-1) + (-2)]
(-3+10) × (-1-2)
7×-3
-21
3x2 – 2xy – 5y2
3 (-1)2 – 2 (-1) (-2) – 5 (-2)2
3 – 4 – 20
– 21
∴ the given expression is verified.
18. (x2y – 1) (3 – 2x2y)
Solution:
Now let us simplify the given expression
(x2y – 1) × (3 – 2x2y)
x2y (3 – 2x2y) – 1 (3 – 2x2y)
3x2y – 2x4y2 – 3 + 2x2y
5x2y – 2x4y2 – 3
Let us substitute the given values x = – 1 and y = – 2, then
(x2y – 1) × (3 – 2x2y)[(-1)2 (-2) – 1] × [3 – 2 (-1)2 (-2)
(-2 – 1) × (3 + 4)
-3 × 7
-21
5x2y – 2x4y2 – 3[-2 (-1)4 (-2)2 + 5 (-1)2 (2) – 3]
– 8 – 10 – 3
-21
∴ the given expression is verified.
19. (1/3x – y2/5) (1/3x + y2/5)
Solution:
Now let us simplify the given expression
(1/3x – y2/5) × (1/3x + y2/5)
(1/3x) 2 – (y2/5)2
(1/3x – y2/5) (1/3x + y2/5)
1/9x2 – 1/25y4
Let us substitute the given values x = – 1 and y = – 2, then
(1/3x – y2/5) × (1/3x + y2/5)
(1/3(-1) – (-2)2/5) × (1/3(-1) + (-2)2/5)
(-17/15) × (7/15)
-119/225
1/9x2 – 1/25y4
1/9 (-1)2 – 1/25 (-2)4
1/9 -16/25
-119/225
∴ the given expression is verified.
Simplify:
20. x2 (x + 2y) (x – 3y)
Solution:
Now let us simplify the given expression
x2 (x + 2y) (x – 3y)
x2 (x2 – 3xy + 2xy – 3y2)
x2 (x2 – xy – 6y2)
x4 – x3y – 6x2y2
21. (x2 – 2y2) (x + 4y)x2y2
Solution:
Now let us simplify the given expression
(x2 – 2y2) (x + 4y)x2y2
(x3 + 4x2y – 2xy2 – 8y3) × x2y2
x5y2 + 4x4y3 – 2x3y4 – 8x2y5
22. a2b2 (a + 2b) (3a + b)
Solution:
Now let us simplify the given expression
a2b2 (a + 2b) (3a + b)
a2b2 (3a2 + ab + 6ab + 2b2)
a2b2 (3a2 + 7ab + 2b2)
3a4b2 + 7a3b3 + 2a2b4
23. x2 (x – y) y2 (x + 2y)
Solution:
Now let us simplify the given expression
x2 (x – y) y2 (x + 2y)
x2y2 (x2 + 2xy – xy – 2y2)
x2y2 (x2 + xy – 2y2)
x4y2 + x3y3 – 2x2y4
24. (x3 – 2x2 + 5x – 7) (2x – 3)
Solution:
Now let us simplify the given expression
(x3 – 2x2 + 5x – 7) (2x – 3)
2x4 – 4x3 + 10x2 – 14x – 3x3 + 6x2 – 15x + 21
2x4 – 7x3 + 16x2 – 29x + 21
25. (5x + 3) (x – 1) (3x – 2)
Solution:
Now let us simplify the given expression
(5x + 3) (x – 1) (3x – 2)
(5x2 – 2x – 3) (3x – 2)
15x3 – 6x2 – 9x – 10x2 + 4x + 6
15x3 – 16x2 – 5x + 6
26. (5 – x) (6 – 5x) (2 – x)
Solution:
Now let us simplify the given expression
(5 – x) (6 – 5x) (2 – x)
(x2 – 7x + 10) (6 – 5x)
-5x3 + 35x2 – 50x + 6x2 – 42x + 60
60 – 92x + 41x2 – 5x3
27. (2x2 + 3x – 5) (3x2 – 5x + 4)
Solution:
Now let us simplify the given expression
(2x2 + 3x – 5) (3x2 – 5x + 4)
6x4 + 9x3 – 15x2 – 10x3 – 15x2 + 25x + 8x2 + 12x – 20
6x4 – x3 – 22x2 + 37x – 20
28. (3x – 2) (2x – 3) + (5x – 3) (x + 1)
Solution:
Now let us simplify the given expression
(3x – 2) (2x – 3) + (5x – 3) (x + 1)
6x2 – 9x – 4x + 6 + 5x2 + 5x – 3x – 3
11x2 – 11x + 3
29. (5x – 3) (x + 2) – (2x + 5) (4x – 3)
Solution:
Now let us simplify the given expression
(5x – 3) (x + 2) – (2x + 5) (4x – 3)
5x2 + 10x – 3x – 6 – 8x2 + 6x – 20x + 15
-3x2 – 7x + 9
30. (3x + 2y) (4x + 3y) – (2x – y) (7x – 3y)
Solution:
Now let us simplify the given expression
(3x + 2y) (4x + 3y) – (2x – y) (7x – 3y)
12x2 + 9xy + 8xy
12x2 + 9xy + 8xy + 6y2 – 14x2 + 6xy + 7xy – 3y2
-2x2 + 3y2 + 30xy
31. (x2 – 3x + 2) (5x – 2) – (3x2 + 4x – 5) (2x – 1)
Solution:
Now let us simplify the given expression
(x2 – 3x + 2) (5x – 2) – (3x2 + 4x – 5) (2x – 1)
5x3 – 15x2 + 10x – 2x2 + 6x – 4 – (6x3 + 8x2 – 10x – 3x2 – 4x + 5)
5x3 – 6x3 – 15x2 – 2x2 – 5x2 + 16x + 14x – 4 – 5
– x3 – 22x2 + 30x – 9
32. (x3 – 2x2 + 3x – 4) (x – 1) – (2x – 3) (x2 – x + 1)
Solution:
Now let us simplify the given expression
(x3 – 2x2 + 3x – 4) (x – 1) – (2x – 3) (x2 – x + 1)
x4 – 2x3 + 3x2 – 4x – x3 + 2x2 – 3x + 4 – (2x3 – 2x2 + 2x – 3x2 + 3x – 3)
x4 – 3x3 + 5x2 – 7x + 4 – 2x3 + 5x2 – 5x + 3
x4 – 5x3 + 10x2 – 12x + 7
EXERCISE 6.6 PAGE NO: 6.43
1. Write the following squares of binomials as trinomials:
(i) (x + 2)2
(ii) (8a + 3b)2
(iii) (2m + 1)2
(iv) (9a + 1/6)2
(v) (x + x2/2)2
(vi) (x/4 – y/3)2
(vii) (3x – 1/3x)2
(viii) (x/y – y/x)2
(ix) (3a/2 – 5b/4)2
(x) (a2b – bc2)2
(xi) (2a/3b + 2b/3a)2
(xii) (x2 – ay)2
Solution:
(i) (x + 2)2
Let us express the given expression in trinomial
x2 + 2 (x) (2) + 22
x2 + 4x + 4
(ii) (8a + 3b)2
Let us express the given expression in trinomial
(8a)2 + 2 (8a) (3b) + (3b)2
64a2 + 48ab + 9b2
(iii) (2m + 1)2
Let us express the given expression in trinomial
(2m)2 + 2 (2m) (1) + 12
4m2 + 4m + 1
(iv) (9a + 1/6)2
Let us express the given expression in trinomial
(9a)2 + 2 (9a) (1/6) + (1/6)2
81a2 + 3a + 1/36
(v) (x + x2/2)2
Let us express the given expression in trinomial
(x)2 + 2 (x) (x2/2) + (x2/2)2
x2 + x3 + 1/4x4
(vi) (x/4 – y/3)2
Let us express the given expression in trinomial
(x/4)2 – 2 (x/4) (y/3) + (y/3)2
1/16x2 – xy/6 + 1/9y2
(vii) (3x – 1/3x)2
Let us express the given expression in trinomial
(3x)2 – 2 (3x) (1/3x) + (1/3x)2
9x2 – 2 + 1/9x2
(viii) (x/y – y/x)2
Let us express the given expression in trinomial
(x/y)2 – 2 (x/y) (y/x) + (y/x)2
x2/y2 – 2 + y2/x2
(ix) (3a/2 – 5b/4)2
Let us express the given expression in trinomial
(3a/2)2 – 2 (3a/2) (5b/4) + (5b/4)2
9/4a2 – 15/4ab + 25/16b2
(x) (a2b – bc2)2
Let us express the given expression in trinomial
(a2b)2 – 2 (a2b) (bc2) + (bc2)2
a4b2 – 2a2b2c2 + b2c4
(xi) (2a/3b + 2b/3a)2
Let us express the given expression in trinomial
(2a/3b)2 + 2 (2a/3b) (2b/3a) + (2b/3a)2
4a2/9b2 + 8/9 + 4b2/9a2
(xii) (x2 – ay)2
Let us express the given expression in trinomial
(x2)2 – 2 (x2) (ay) + (ay)2
x4 – 2x2ay + a2y2
2. Find the product of the following binomials:
(i) (2x + y) (2x + y)
(ii) (a + 2b) (a – 2b)
(iii) (a2 + bc) (a2 – bc)
(iv) (4x/5 – 3y/4) (4x/5 + 3y/4)
(v) (2x + 3/y) (2x – 3/y)
(vi) (2a3 + b3) (2a3 – b3)
(vii) (x4 + 2/x2) (x4 – 2/x2)
(viii) (x3 + 1/x3) (x3 – 1/x3)
Solution:
(i) (2x + y) (2x + y)
Let us find the product of the given expression
2x (2x + y) + y (2x + y)
4x2 + 2xy + 2xy + y2
4x2 + 4xy + y2
(ii) (a + 2b) (a – 2b)
Let us find the product of the given expression
a (a – 2b) + 2b (a – 2b)
a2 – 2ab + 2ab – 4b2
a2 – 4b2
(iii) (a2 + bc) (a2 – bc)
Let us find the product of the given expression
a2 (a2 – bc) + bc (a2 – bc)
a4 – a2bc + bca2 – b2c2
a4 – b2c2
(iv) (4x/5 – 3y/4) (4x/5 + 3y/4)
Let us find the product of the given expression
4x/5 (4x/5 + 3y/4) – 3y/4 (4x/5 + 3y/4)
16/25x2 + 12/20yx – 12/20xy – 9y2/16
16/25x2 – 9/16y2
(v) (2x + 3/y) (2x – 3/y)
Let us find the product of the given expression
2x (2x – 3/y) + 3/y (2x – 3/y)
4x2 – 6x/y + 6x/y – 9/y2
4x2 – 9/y2
(vi) (2a3 + b3) (2a3 – b3)
Let us find the product of the given expression
2a3 (2a3 – b3) + b3 (2a3 – b3)
4a6 – 2a3b3 + 2a3b3 – b6
4a6 – b6
(vii) (x4 + 2/x2) (x4 – 2/x2)
Let us find the product of the given expression
x4 (x4 – 2/x2) + 2/x2 (x4 – 2/x2)
x8 – 2x2 + 2x2 – 4/x4
(x8 – 4/x4)
(viii) (x3 + 1/x3) (x3 – 1/x3)
Let us find the product of the given expression
x3 (x3 – 1/x3) + 1/x3 (x3 – 1/x3)
x6 – 1 + 1 – 1/x6
x6 – 1/x6
3. Using the formula for squaring a binomial, evaluate the following:
(i) (102)2
(ii) (99)2
(iii) (1001)2
(iv) (999)2
(v) (703)2
Solution:
(i) (102)2
We can express 102 as 100 + 2
So, (102)2 = (100 + 2)2
Upon simplification we get,
(100 + 2)2 = (100)2 + 2 (100) (2) + 22
= 10000 + 400 + 4
= 10404
(ii) (99)2
We can express 99 as 100 – 1
So, (99)2 = (100 – 1)2
Upon simplification we get,
(100 – 1)2 = (100)2 – 2 (100) (1) + 12
= 10000 – 200 + 1
= 9801
(iii) (1001)2
We can express 1001 as 1000 + 1
So, (1001)2 = (1000 + 1)2
Upon simplification we get,
(1000 + 1)2 = (1000)2 + 2 (1000) (1) + 12
= 1000000 + 2000 + 1
= 1002001
(iv) (999)2
We can express 999 as 1000 – 1
So, (999)2 = (1000 – 1)2
Upon simplification we get,
(1000 – 1)2 = (1000)2 – 2 (1000) (1) + 12
= 1000000 – 2000 + 1
= 998001
(v) (703)2
We can express 700 as 700 + 3
So, (703)2 = (700 + 3)2
Upon simplification we get,
(700 + 3)2 = (700)2 + 2 (700) (3) + 32
= 490000 + 4200 + 9
= 494209
4. Simplify the following using the formula: (a – b) (a + b) = a2 – b2 :
(i) (82)2 – (18)2
(ii) (467)2 – (33)2
(iii) (79)2 – (69)2
(iv) 197 × 203
(v) 113 × 87
(vi) 95 × 105
(vii) 1.8 × 2.2
(viii) 9.8 × 10.2
Solution:
(i) (82)2 – (18)2
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
(82)2 – (18)2 = (82 – 18) (82 + 18)
= 64 × 100
= 6400
(ii) (467)2 – (33)2
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
(467)2 – (33)2 = (467 – 33) (467 + 33)
= (434) (500)
= 217000
(iii) (79)2 – (69)2
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
(79)2 – (69)2 = (79 + 69) (79 – 69)
= (148) (10)
= 1480
(iv) 197 × 203
We can express 203 as 200 + 3 and 197 as 200 – 3
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
197 × 203 = (200 – 3) (200 + 3)
= (200)2 – (3)2
= 40000 – 9
= 39991
(v) 113 × 87
We can express 113 as 100 + 13 and 87 as 100 – 13
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
113 × 87 = (100 – 13) (100 + 13)
= (100)2 – (13)2
= 10000 – 169
= 9831
(vi) 95 × 105
We can express 95 as 100 – 5 and 105 as 100 + 5
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
95 × 105 = (100 – 5) (100 + 5)
= (100)2 – (5)2
= 10000 – 25
= 9975
(vii) 1.8 × 2.2
We can express 1.8 as 2 – 0.2 and 2.2 as 2 + 0.2
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
1.8 × 2.2 = (2 – 0.2) ( 2 + 0.2)
= (2)2 – (0.2)2
= 4 – 0.04
= 3.96
(viii) 9.8 × 10.2
We can express 9.8 as 10 – 0.2 and 10.2 as 10 + 0.2
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
9.8 × 10.2 = (10 – 0.2) (10 + 0.2)
= (10)2 – (0.2)2
= 100 – 0.04
= 99.96
5. Simplify the following using the identities:
(i) ((58)2 – (42)2)/16
(ii) 178 × 178 – 22 × 22
(iii) (198 × 198 – 102 × 102)/96
(iv) 1.73 × 1.73 – 0.27 × 0.27
(v) (8.63 × 8.63 – 1.37 × 1.37)/0.726
Solution:
(i) ((58)2 – (42)2)/16
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
((58)2 – (42)2)/16 = ((58-42) (58+42)/16)
= ((16) (100)/16)
= 100
(ii) 178 × 178 – 22 × 22
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
178 × 178 – 22 × 22 = (178)2 – (22)2
= (178-22) (178+22)
= 200 × 156
= 31200
(iii) (198 × 198 – 102 × 102)/96
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
(198 × 198 – 102 × 102)/96 = ((198)2 – (102)2)/96
= ((198-102) (198+102))/96
= (96 × 300)/96
= 300
(iv) 1.73 × 1.73 – 0.27 × 0.27
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
1.73 × 1.73 – 0.27 × 0.27 = (1.73)2 – (0.27)2
= (1.73-0.27) (1.73+0.27)
= 1.46 × 2
= 2.92
(v) (8.63 × 8.63 – 1.37 × 1.37)/0.726
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
(8.63 × 8.63 – 1.37 × 1.37)/0.726 = ((8.63)2 – (1.37)2)/0.726
= ((8.63-1.37) (8.63+1.37))/0.726
= (7.26 × 10)/0.726
= 72.6/0.726
= 100
6. Find the value of x, if:
(i) 4x = (52)2 – (48)2
(ii) 14x = (47)2 – (33)2
(iii) 5x = (50)2 – (40)2
Solution:
(i) 4x = (52)2 – (48)2
Let us simplify to find the value of x by using the formula (a – b) (a + b) = a2 – b2
4x = (52)2 – (48)2
4x = (52 – 48) (52 + 48)
4x = 4 × 100
4x = 400
x = 100
(ii) 14x = (47)2 – (33)2
Let us simplify to find the value of x by using the formula (a – b) (a + b) = a2 – b2
14x = (47)2 – (33)2
14x = (47 – 33) (47 + 33)
14x = 14 × 80
x = 80
(iii) 5x = (50)2 – (40)2
Let us simplify to find the value of x by using the formula (a – b) (a + b) = a2 – b2
5x = (50)2 – (40)2
5x = (50 – 40) (50 + 40)
5x = 10 × 90
5x = 900
x = 180
7. If x + 1/x =20, find the value of x2 + 1/ x2.
Solution:
We know that x + 1/x = 20
So when squaring both sides, we get
(x + 1/x)2 = (20)2
x2 + 2 × x × 1/x + (1/x)2 = 400
x2 + 2 + 1/x2 = 400
x2 + 1/x2 = 398
8. If x – 1/x = 3, find the values of x2 + 1/ x2 and x4 + 1/ x4.
Solution:
We know that x – 1/x = 3
So when squaring both sides, we get
(x – 1/x)2 = (3)2
x2 – 2 × x × 1/x + (1/x)2 = 9
x2 – 2 + 1/x2 = 9
x2 + 1/x2 = 9+2
x2 + 1/x2 = 11
Now again when we square on both sides we get,
(x2 + 1/x2)2 = (11)2
x4 + 2 × x2 × 1/x2 + (1/x2)2 = 121
x4 + 2 + 1/x4 = 121
x4 + 1/x4 = 121-2
x4 + 1/x4 = 119
∴ x2 + 1/x2 = 11
x4 + 1/x4 = 119
9. If x2 + 1/x2 = 18, find the values of x + 1/ x and x – 1/ x.
Solution:
We know that x2 + 1/x2 = 18
When adding 2 on both sides, we get
x2 + 1/x2 + 2 = 18 + 2
x2 + 1/x2 + 2 × x × 1/x = 20
(x + 1/x)2 = 20
x + 1/x = √20
When subtracting 2 from both sides, we get
x2 + 1/x2 – 2 × x × 1/x = 18 – 2
(x – 1/x)2 = 16
x – 1/x = √16
x – 1/x = 4
10. If x + y = 4 and xy = 2, find the value of x2 + y2
Solution:
We know that x + y = 4 and xy = 2
Upon squaring on both sides of the given expression, we get
(x + y)2 = 42
x2 + y2 + 2xy = 16
x2 + y2 + 2 (2) = 16 (since xy=2)
x2 + y2 + 4 = 16
x2 + y2 = 16 – 4
x2 + y2 =12
11. If x – y = 7 and xy = 9, find the value of x2+y2
Solution:
We know that x – y = 7 and xy = 9
Upon squaring on both sides of the given expression, we get
(x – y)2 = 72
x2 + y2 – 2xy = 49
x2 + y2 – 2 (9) = 49 (since xy=9)
x2 + y2 – 18 = 49
x2 + y2 = 49 + 18
x2 + y2 =67
12. If 3x + 5y = 11 and xy = 2, find the value of 9x2 + 25y2
Solution:
We know that 3x + 5y = 11 and xy = 2
Upon squaring on both sides of the given expression, we get
(3x + 5y)2 = 112
(3x)2 + (5y)2 + 2(3x)(5y) = 121
9x2 + 25y2 + 2 (15xy) = 121 (since xy=2)
9x2 + 25y2 + 2(15(2)) = 121
9x2 + 25y2 + 60 = 121
9x2 + 25y2 = 121-60
9x2 + 25y2 = 61
13. Find the values of the following expressions:
(i) 16x2 + 24x + 9 when x = 7/4
(ii) 64x2 + 81y2 + 144xy when x = 11 and y = 4/3
(iii) 81x2 + 16y2 – 72xy when x = 2/3 and y = ¾
Solution:
(i) 16x2 + 24x + 9 when x = 7/4
Let us find the values using the formula (a + b)2 = a2 + b2 + 2ab
(4x)2 + 2 (4x) (3) + 32
(4x + 3)2
Evaluating when x = 7/4[4 (7/4) + 3]2
(7 + 3)2
100
(ii) 64x2 + 81y2 + 144xy when x = 11 and y = 4/3
Let us find the values using the formula (a + b)2 = a2 + b2 + 2ab
(8x)2 + 2 (8x) (9y) + (9y)2 (8x + 9y)
Evaluating when x = 11 and y = 4/3[8 (11) + 9 (4/3)]2
(88 + 12)2
(100)2
10000
(iii) 81x2 + 16y2 – 72xy when x = 2/3 and y = ¾
Let us find the values using the formula (a + b)2 = a2 + b2 + 2ab
(9x)2 + (4y)2 – 2 (9x) (4y)
(9x – 4y)2
Putting x = 2/3 and y = 3/4[9 (2/3) – 4 (3/4)]2
(6 – 3)2
32
9
14. If x + 1/x = 9 find the value of x4 + 1/ x4.
Solution:
We know that x + 1/x = 9
So when squaring both sides, we get
(x + 1/x)2 = (9)2
x2 + 2 × x × 1/x + (1/x)2 = 81
x2 + 2 + 1/x2 = 81
x2 + 1/x2 = 81 – 2
x2 + 1/x2 = 79
Now again when we square on both sides we get,
(x2 + 1/x2)2 = (79)2
x4 + 2 × x2 × 1/x2 + (1/x2)2 = 6241
x4 + 2 + 1/x4 = 6241
x4 + 1/x4 = 6241- 2
x4 + 1/x4 = 6239
∴ x4 – 1/x4 = 6239
15. If x + 1/x = 12 find the value of x – 1/x.
Solution:
We know that x + 1/x = 12
So when squaring both sides, we get
(x + 1/x)2 = (12)2
x2 + 2 × x × 1/x + (1/x)2 = 144
x2 + 2 + 1/x2 = 144
x2 + 1/x2 = 144 – 2
x2 + 1/x2 = 142
When subtracting 2 from both sides, we get
x2 + 1/x2 – 2 × x × 1/x = 142 – 2
(x – 1/x)2 = 140
x – 1/x = √140
16. If 2x + 3y = 14 and 2x – 3y = 2, find value of xy. [Hint: Use (2x+3y) 2 – (2x-3y) 2 = 24xy]
Solution:
We know that the given equations are
2x + 3y = 14… equation (1)
2x – 3y = 2… equation (2)
Now, let us square both the equations and subtract equation (2) from equation (1), we get,
(2x + 3y) 2 – (2x – 3y) 2 = (14)2 – (2)2
4x2 + 9y2 + 12xy – 4x2 – 9y2 + 12xy = 196 – 4
24 xy = 192
xy = 8
∴ the value of xy is 8.
17. If x2 + y2 = 29 and xy = 2, find the value of
(i) x + y
(ii) x – y
(iii) x4 + y4
Solution:
(i) x + y
We know that
x2 + y2 = 29
x2 + y2 + 2xy – 2xy = 29
(x + y) 2 – 2 (2) = 29
(x + y) 2 = 29 + 4
x + y = ± √33
(ii) x – y
We know that
x2 + y2 = 29
x2 + y2 + 2xy – 2xy = 29
(x – y)2 + 2 (2) = 29
(x – y)2 + 4 = 29
(x – y)2 = 25
(x – y) = ± 5
(iii) x4 + y4
We know that
x2 + y2 = 29
Squaring both sides, we get
(x2 + y2)2 = (29)2
x4 + y4 + 2x2y2 = 841
x4 + y4 + 2 (2)2 = 841
x4 + y4 = 841 – 8
x4 + y4 = 833
18. What must be added each of the following expression to make it a whole square?
(i) 4x2 – 12x + 7
(ii) 4x2 – 20x + 20
Solution:
(i) 4x2 – 12x + 7
(2x)2 – 2 (2x) (3) + 32 – 32 + 7
(2x – 3)2 – 9 + 7
(2x – 3)2 – 2
∴ 2 must be added to the expression to make it a whole square.
(ii) 4x2 – 20x + 20
(2x)2 – 2 (2x) (5) + 52 – 52 + 20
(2x – 5)2 – 25 + 20
(2x – 5)2 – 5
∴ 5 must be added to the expression to make it a whole square.
19. Simplify:
(i) (x – y) (x + y) (x2 + y2) (x4 + y4)
(ii) (2x – 1) (2x + 1) (4x2 + 1) (16x4 + 1)
(iii) (7m – 8n) 2 + (7m + 8n) 2
(iv) (2.5p – 1.5q) 2 – (1.5p – 2.5q) 2
(v) (m2 – n2m) 2 + 2m3n2
Solution:
(i) (x – y) (x + y) (x2 + y2) (x4 + y4)
B7 grouping the values
(x2 – y2) (x2 + y2) (x4 + y4)[(x2)2 – (y2)2] (x4 + y4)
(x4 – y4) (x4 – y4)[(x4)2 – (y4)2]
x8 – y8
(ii) (2x – 1) (2x + 1) (4x2 + 1) (16x4 + 1)
Let us simplify the expression by grouping[(2x)2 – (1)2] (4x2 + 1) (16x4 + 1)
(4x2 – 1) (4x2 + 1) (16x4 + 1) 1[(4x2)2 – (1)2] (16x4 + 1) 1
(16x4 – 1) (16x4 + 1) 1[(16x4)2 – (1)2] 1
256x8 – 1
(iii) (7m – 8n)2 + (7m + 8n)2
Upon expansion
(7m)2 + (8n)2 – 2(7m)(8n) + (7m)2 + (8n)2 + 2(7m)(8n)
(7m)2 + (8n)2 – 112mn + (7m)2 + (8n)2 + 112mn
49m2 + 64n2 + 49m2 + 64n2
By grouping the similar expression we get,
98m2 + 64n2 + 64n2
98m2 + 128n2
(iv) (2.5p – 1.5q)2 – (1.5p – 2.5q)2
Upon expansion
(2.5p)2 + (1.5q)2 – 2 (2.5p) (1.5q) – (1.5p)2 – (2.5q)2 + 2 (1.5p) (2.5q)
6.25p2 + 2.25q2 – 2.25p2 – 6.25q2
By grouping the similar expression we get,
4p2 – 6.25q2 + 2.25q2
4p2 – 4q2
4 (p2 – q2)
(v) (m2 – n2m)2 + 2m3n2
Upon expansion using (a + b) 2 formula
(m2)2 – 2 (m2) (n2) (m) + (n2m) 2 + 2m3n2
m4 – 2m3n2 + (n2m)2 + 2m3n2
m4+ n4m2 – 2m3n2 + 2m3n2
m4+ m2n4
20. Show that:
(i) (3x + 7)2 – 84x = (3x – 7)2
(ii) (9a – 5b)2 + 180ab = (9a + 5b)2
(iii) (4m/3 – 3n/4)2 + 2mn = 16m2/9 + 9n2/16
(iv) (4pq + 3q)2 – (4pq – 3q)2 = 48pq2
(v) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
Solution:
(i) (3x + 7)2 – 84x = (3x – 7)2
Let us consider LHS (3x + 7)2 – 84x
By using the formula (a + b)2 = a2 + b2 + 2ab
(3x)2 + (7)2 + 2 (3x) (7) – 84x
(3x)2 + (7)2 + 42x – 84x
(3x)2 + (7)2 – 42x
(3x)2 + (7)2 – 2 (3x) (7)
(3x – 7)2 = R.H.S
Hence, proved
(ii) (9a – 5b)2 + 180ab = (9a + 5b)2
Let us consider LHS (9a – 5b)2 + 180ab
By using the formula (a + b)2 = a2 + b2 + 2ab
(9a)2 + (5b)2 – 2 (9a) (5b) + 180ab
(9a)2 6 (5b)2 – 90ab + 180ab
(9a)2 + (5b)2 + 9ab
(9a)2 + (5b)2 + 2 (9a) (5b)
(9a + 5b)2 = R.H.S
Hence, proved
(iii) (4m/3 – 3n/4)2 + 2mn = 16m2/9 + 9n2/16
Let us consider LHS (4m/3 – 3n/4)2 + 2mn
(4m/3)2 + (3n/4)2 – 2mn + 2mn
(4m/3)2 + (3n/4)2
16/9m2 + 9/16n2 = R.H.S
Hence, proved
(iv) (4pq + 3q)2 – (4pq – 3q)2 = 48pq2
Let us consider LHS (4pq + 3q)2 – (4pq – 3q)2
(4pq)2 + (3q)2 + 2 (4pq) (3q) – (4pq)2 – (3q)2 + 2(4pq)(3q)
24pq2 + 24pq2
48pq2 = RHS
Hence, proved
(v) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
Let us consider LHS (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a)
By using the identity (a – b) (a + b) = a2 – b2
We get,
(a2 – b2) + (b2 – c2) + (c2 – a2)
a2 – b2 + b2 – c2 + c2 – a2
0 = R.H.S
Hence, proved
EXERCISE 6.7 PAGE NO: 6.47
1. Find the following products:
(i) (x + 4) (x + 7)
(ii) (x – 11) (x + 4)
(iii) (x + 7) (x – 5)
(iv) (x – 3) (x – 2)
(v) (y2 – 4) (y2 – 3)
(vi) (x + 4/3) (x + 3/4)
(vii) (3x + 5) (3x + 11)
(viii) (2x2 – 3) (2x2 + 5)
(ix) (z2 + 2) (z2 – 3)
(x) (3x – 4y) (2x – 4y)
(xi) (3x2 – 4xy) (3x2 – 3xy)
(xii) (x + 1/5) (x + 5)
(xiii) (z + 3/4) (z + 4/3)
(xiv) (x2 + 4) (x2 + 9)
(xv) (y2 + 12) (y2 + 6)
(xvi) (y2 + 5/7) (y2 – 14/5)
(xvii) (p2 + 16) (p2 – 1/4)
Solution:
(i) (x + 4) (x + 7)
Let us simplify the given expression
x (x + 7) + 4 (x + 7)
x2 + 7x + 4x + 28
x2 + 11x + 28
(ii) (x – 11) (x + 4)
Let us simplify the given expression
x (x + 4) – 11 (x + 4)
x2 + 4x – 11x – 44
x2 – 7x – 44
(iii) (x + 7) (x – 5)
Let us simplify the given expression
x (x – 5) + 7 (x – 5)
x2 – 5x + 7x – 35
x2 + 2x – 35
(iv) (x – 3) (x – 2)
Let us simplify the given expression
x (x – 2) – 3 (x – 2)
x2 – 2x – 3x + 6
x2 – 5x + 6
(v) (y2 – 4) (y2 – 3)
Let us simplify the given expression
y2 (y2 – 3) – 4 (y2 – 3)
y4 – 3y2 – 4y2 + 12
y4 – 7y2 + 12
(vi) (x + 4/3) (x + 3/4)
Let us simplify the given expression
x (x + 3/4) + 4/3 (x + 3/4)
x2 + 3x/4 + 4x/3 + 12/12
x2 + 3x/4 + 4x/3 + 1
x2 + 25x/12 + 1
(vii) (3x + 5) (3x + 11)
Let us simplify the given expression
3x (3x + 11) + 5 (3x + 11)
9x2 + 33x + 15x + 55
9x2 + 48x + 55
(viii) (2x2 – 3) (2x2 + 5)
Let us simplify the given expression
2x2 (2x2 + 5) – 3 (2x2 + 5)
4x4 + 10x2 – 6x2 – 15
4x4 + 4x2 – 15
(ix) (z2 + 2) (z2 – 3)
Let us simplify the given expression
z2 (z2 – 3) + 2 (z2 – 3)
z4 – 3z2 + 2z2 – 6
z4 – z2 – 6
(x) (3x – 4y) (2x – 4y)
Let us simplify the given expression
3x (2x – 4y) – 4y (2x – 4y)
6x2 – 12xy – 8xy + 16y2
6x2 – 20xy + 16y2
(xi) (3x2 – 4xy) (3x2 – 3xy)
Let us simplify the given expression
3x2 (3x2 – 3xy) – 4xy (3x2 – 3xy)
9x4 – 9x3y – 12x3y + 12x2y2
9x4 – 21x3y + 12x2y2
(xii) (x + 1/5) (x + 5)
Let us simplify the given expression
x (x + 1/5) + 5 (x + 1/5)
x2 + x/5 + 5x + 1
x2 + 26/5x + 1
(xiii) (z + 3/4) (z + 4/3)
Let us simplify the given expression
z (z + 4/3) + 3/4 (z + 4/3)
z2 + 4/3z + 3/4z + 12/12
z2 + 4/3z + 3/4z + 1
z2 + 25/12z + 1
(xiv) (x2 + 4) (x2 + 9)
Let us simplify the given expression
x2 (x2 + 9) + 4 (x2 + 9)
x4 + 9x2 + 4x2 + 36
x4 + 13x2 + 36
(xv) (y2 + 12) (y2 + 6)
Let us simplify the given expression
y2 (y2 + 6) + 12 (y2 + 6)
y4 + 6y2 + 12y2 + 72
y4 + 18y2 + 72
(xvi) (y2 + 5/7) (y2 – 14/5)
Let us simplify the given expression
y2 (y2 – 14/5) + 5/7 (y2 – 14/5)
y4 – 14/5y2 + 5/7y2 – 2
y4 – 73/35y2 – 2
(xvii) (p2 + 16) (p2 – 1/4)
Let us simplify the given expression
p2 (p2 – 1/4) + 16 (p2 – 1/4)
p4 – 1/4p2 + 16p2 – 4
p4 + 63/4p2 – 4
2. Evaluate the following:
(i) 102 × 106
(ii) 109 × 107
(iii) 35 × 37
(iv) 53 × 55
(v) 103 × 96
(vi) 34 × 36
(vii) 994 × 1006
Solution:
(i) 102 × 106
We can express 102 as 100 + 2 and 106 as 100 + 6
Now let us simplify
102 × 106 = (100 + 2) (100 + 6)
= 100 (100 + 6) + 2 (100 + 6)
= 10000 + 600 + 200 + 12
= 10812
(ii) 109 × 107
We can express 109 as 100 + 9 and 107 as 100 + 7
Now let us simplify
109 × 107 = (100 + 9) (100 + 7)
= 100 (100 + 7) + 9 (100 + 7)
= 10000 + 700 + 900 + 63
= 11663
(iii) 35 × 37
We can express 35 as 30 + 5 and 37 as 30 + 7
Now let us simplify
35 × 37 = (30 + 5) (30 + 7)
= 30 (30 + 7) + 5 (30 + 7)
= 900 + 210 + 150 + 35
= 1295
(iv) 53 × 55
We can express 53 as 50 + 3 and 55 as 50 + 5
Now let us simplify
53 × 55 = (50 + 3) (50 + 5)
= 50 (50 + 5) + 3 (50 + 5)
= 2500 + 250 + 150 + 15
= 2915
(v) 103 × 96
We can express 103 as 100 + 3 and 96 as 100 – 4
Now let us simplify
103 × 96 = (100 + 3) (100 – 4)
= 100 (100 – 4) + 3 (100 – 4)
= 10000 – 400 + 300 – 12
= 10000 – 112
= 9888
(vi) 34 × 36
We can express 34 as 30 + 4 and 36 as 30 + 6
Now let us simplify
34 × 36 = (30 + 4) (30 + 6)
= 30 (30 + 6) + 4 (30 + 6)
= 900 + 180 + 120 + 24
= 1224
(vii) 994 × 1006
We can express 994 as 1000 – 6 and 1006 as 1000 + 6
Now let us simplify
994 × 1006 = (1000 – 6) (1000 + 6)
= 1000 (1000 + 6) – 6 (1000 + 6)
= 1000000 + 6000 – 6000 – 36
= 999964
RD Sharma Solutions for Class 8 Maths Chapter 6: Download PDF
RD Sharma Solutions for Class 8 Maths Chapter 6–Algebraic Expressions and Identities
Chapterwise RD Sharma Solutions for Class 8 Maths :
- Chapter 1–Rational Numbers
- Chapter 2–Powers
- Chapter 3–Squares and Square Roots
- Chapter 4–Cubes and Cube Roots
- Chapter 5–Playing with Numbers
- Chapter 6–Algebraic Expressions and Identities
- Chapter 7–Factorization
- Chapter 8–Division of Algebraic Expressions
- Chapter 9–Linear Equation in One Variable
- Chapter 10–Direct and Inverse Variations
- Chapter 11–Time and Work
- Chapter 12–Percentage
- Chapter 13–Profit, Loss, Discount and Value Added Tax (VAT)
- Chapter 14–Compound Interest
- Chapter 15–Understanding Shapes- I (Polygons)
- Chapter 16–Understanding Shapes- II (Quadrilaterals)
- Chapter 17–Understanding Shapes- III (Special Types of Quadrilaterals)
- Chapter 18–Practical Geometry (Constructions)
- Chapter 19–Visualising Shapes
- Chapter 20–Mensuration – I (Area of a Trapezium and a Polygon)
- Chapter 21–Mensuration – II (Volumes and Surface Areas of a Cuboid and a cube)
- Chapter 22–Mensuration – III (Surface Area and Volume of a Right Circular Cylinder)
- Chapter 23–Data Handling – I (Classification and Tabulation of Data)
- Chapter 24–Data Handling – II (Graphical Representation of Data as Histogram)
- Chapter 25–Data Handling – III (Pictorial Representation of Data as Pie Charts or Circle Graphs)
- Chapter 26–Data Handling – IV (Probability)
- Chapter 27–Introduction to Graphs
About RD Sharma
RD Sharma isn’t the kind of author you’d bump into at lit fests. But his bestselling books have helped many CBSE students lose their dread of maths. Sunday Times profiles the tutor turned internet star
He dreams of algorithms that would give most people nightmares. And, spends every waking hour thinking of ways to explain concepts like ‘series solution of linear differential equations’. Meet Dr Ravi Dutt Sharma — mathematics teacher and author of 25 reference books — whose name evokes as much awe as the subject he teaches. And though students have used his thick tomes for the last 31 years to ace the dreaded maths exam, it’s only recently that a spoof video turned the tutor into a YouTube star.
R D Sharma had a good laugh but said he shared little with his on-screen persona except for the love for maths. “I like to spend all my time thinking and writing about maths problems. I find it relaxing,” he says. When he is not writing books explaining mathematical concepts for classes 6 to 12 and engineering students, Sharma is busy dispensing his duty as vice-principal and head of department of science and humanities at Delhi government’s Guru Nanak Dev Institute of Technology.