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Published on: 26/09/2019
Algebraic Expressions and Identities
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1.
Find the volume of each rectangular box with given length, breadth and height
| Length | Breadth | Height | |
| (i) | 2ax | 3by | 5cz |
| (ii) | m2n | n2p | p2m |
| (iii) | 2q | 4q2 | 8q3 |
2.
Find the product of 2x, 3ax2 and -5pqx3
3.
The length and breadth of a rectangle are 3x2- 2 and 2x + 5 respectively. Find its area
4.
Verify the identity (x + a)( x + b) =.x2+ (a + b)x + ab for a = 2, b = 3 and x = 4,
5.
Show that (a - b)(a + b) + (b - c)(b + c) + (c - a)(c + a) = 0
6.
Show that (4pq + 3q)2 - (4pq - 3q)2 = 48pq2
7.
Subtract 3xy+ 5yz- 7zx from 5xy- 2yz - 2zx + 10xyz.
8.
Simplify:(2.5p- 1.5q)2 - (1.5p- 2.5q)2
9.
Simplify:(4m + 5n)2 + (5m+ 4n)2
10.
Simplify:(7m - 8n)2 + (7m + 8n)2
11.
Subtract 3l (l- 4m + 5n) from 41 (10n- 3m + 21)
12.
Add 2x (z- x- y) and 2y (z- y- x).
13.
Subtract the sum of 3a - 4b +c and 1- 2a + b from the sum of 3b+c + b and a -2b + 3
14.
Subtract 1+ a2 + b2 from the sum of 1- a2-b2 and a2-b2
15.
Subtract 4a - 7ab + 3b + 12 from 12a- 9ab+ 5b- 3.
1.
(i) 30 abcxyz
(ii) m3n3p3
(iii) 64q6
2.
-30 apqx6
3.
Here, Length = 3x2 - 2
Breadth = 2x + 5
∴ Area= (Length)\(\times\)(Breadth
= (3x2- 2)\(\times\)(2x + 5)
= 3x2(2x + 5) + (-2)(2x + 5)
= (3x2\(\times\)2x) + (5\(\times\)3x2) + [(-2)\(\times\)2x + (-2)\(\times\)5]
= 6x3+ 15x2+ (-4x) + (-10)
= 6x3 + 15x2 - 4x -10
Thus, the required area of the rectangle is 6x3 + 15x2- 4x - 10 sq. units
4.
We have
(x + a)(x + b) = x2+ (a + b)x + ab
Puting a = 2, b = 3 and x = 4, we have
LHS = (x + a)(x + b)
= (4 + 2)(4 + 3)
= 6\(\times\)7 = 42
RHS = x2+ (a + b)x + ab
= (4)2 + (2 + 3)4 + (2\(\times\)3)
= 16 + 5\(\times\)4 + 6
= 16 + 20 + 6 = 42
i.e., LHS = RHS
Thus, the given identity is true for the given values
5.
LHS= (a - b)(a + b) + (b - c)(b + c) + (c - a)(c + a)
= (a2-b2) + (b2-c2) + (c2-a2)
= a2-b2 + b2-c2 +c2-a2
= a2- a2 + b2 - b2 + c2 - c2 = 0 = RHS
Since, LHS = RHS
∴ (a - b)(a + b) + (b - c)(b + c) + (c - a)(c + a) = 0
6.
LHS=(4pq + 3q)2 - (4pq - 3q)2
= [(4pq)2 + 2(4pq)(3q) + (3q)2] - [(4pq)2 - 2(4pq)(3q) + (3q)2]
= 16p2q2 + 24pq2 + 9q2-[16p2q2-24pq2+ 9q2]
= 16p2q2 + 24pq2 + 9q2-16p2q2-24pq2- 9q2
= (16 - 16)p2q2 + (24 + 24)pq2 + (9 - 9)q2
= (0)p2q2 + 48pq2 + (0)q2 = 48pq2 = RHS
Since, LHS = RHS
∴ (4pq + 3q)2 - (4pq - 3q)2 = 48pq2
7.
For subtraction, write the like terms one below the other, we get
5xy - 2yz - 2zx + 10xyz
+ 3xy + 5y z- 7zx
(-) (-) (+)
_______________________
2xy-7yz + 5zx + 10xyz
_______________________
Thus, the required answer is 2xy - 7yz + 5zx + 10xyz
8.
=[(2.5p)2 - 2(2.5p)(1.5q) + (1.5q)2] - [(1.5p)2 - 2(1.5p)(2.5q) + (2.5q)2]
= 6.25p2 - 7.5pq + 2.25q2 - [2.25p2 - 7.5pq + 6.25q2]
= 6.25p2 - 7.5pq + 2.25q2 - 2.25p2 + 7.5pq - 6.25q2
= (6.25 - 2.25)p2 + (-7.5+7.5)pq + (2.25-6.25)q2
= 4p2 + 0(pq) - 4q2 = 4p2 -4q2
9.
[(4m)2 + 2(4m)(5n) + (5n)2] + [(5m)2 + 2(5m)(4n) +(4n)2]
= 16m2 + 40mn + 25n2 + 25m2 + 40mn + 16n2
= (16 + 25)m2+ (40 + 40)mn + (25 + 16)2
= 41m2 + 80mn + 41n2
10.
(7m - 8n)2 + (7m + 8n)2
= [(7m)2 - 2(7m)(8n) + (8n)2] + [(7m)2 + 2(7m)(8n) + (8n)2]
and (a-b)2= a2 -2ab+b2
= (49m2 -112mn + 64n2)+(49m2 +112mn + 64n2)
= 49m2 -112mn + 64n2 + 49m2 +112mn + 64n2
= 98m2 + 128n2
11.
First expression = 3l(l - 4m + 5n)
= (3l) x (l) - (3l) x (4m) + (3l) x (5n)
= 3l2 -12lm + 15ln
Second expression = 4l(10n - 3m + 2l)
= (4l)x (I0n) - (4l) x (3m) + (4l) x (2l)
= 40ln -12lm + 8l2
On subtracting first expression from second expression, we get
40ln -12lm + 8l2
15ln - 12lm + 3l2
(-) (+) (-)
___________________
25ln + 0 + 5l2
___________________
= 25ln + 5l2
12.
First expression = 2x(z - x - y)
= (2x) x z - (2x) x (x) - (2x) xy
= 2xz - 2x2 - 2xy
Second expression = 2y(z - y - x)
= (2y)x(z)-(2y) x (y)-(2y)x(x)
= 2yz - 2y2 - 2yx
On adding above expressions, we get
2xz - 2x2 - 2xy
+ -2yx + 2yz - 2y2
_________________________
2xz - 2x2 - 4xy + 2yz - 2y2
__________________________
13.
4b+2
14.
-a2-3b2
15.
For subtraction, write the like terms one below the other, we get
12a -9ab +5b -3
+4a - 7ab + 3b + 12
(-) (+) (-) (-)
_________________
8a - 2ab + 2b -15
_________________
Thus, the required answer is 8a - 2ab + 2b - 15.
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