(i) 2x-7 = 2(-1)-7 [Putting x =-1] = -2-7 = -9 (ii) -x + 2 = -(-1)+2 [Putting x =-1] = 1 + 2 = 3 Class 7 Maths Chapter 12 Exercise 12.3 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-12/
(i) 2x-7 = 2(-1)-7 [Putting x =-1]
= -2-7 = -9
(ii) -x + 2 = -(-1)+2 [Putting x =-1]
= 1 + 2 = 3
(i) m - 2 = 2-2 [Putting m = 2 ] = 0 (ii) 3m-5 = 3x2-5 [Putting m = 2 ] = 6 – 5 = 1 Class 7 Maths Chapter 12 Exercise 12.3 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-12/
(i) m – 2 = 2-2 [Putting m = 2 ]
= 0
(ii) 3m-5 = 3×2-5 [Putting m = 2 ]
= 6 – 5 = 1
From the sum of (3x – y + 11) + (– y – 11) - (3x – y – 11) = 3x - y +11-y-11-3x+y+11 = 3x - 3x - y - y - y + 11 + 11 - 11 = (3-3)x - (1+111) y 111111 = 0x - y + 11 = -y+11 Class 7 Maths Chapter 12 Exercise 12.2 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/Read more
From the sum of
(3x – y + 11) + (– y – 11) – (3x – y – 11) = 3x – y +11-y-11-3x+y+11
= 3x – 3x – y – y – y + 11 + 11 – 11
= (3-3)x – (1+111) y 111111
= 0x – y + 11 = -y+11
If a = 2,b= -2, find the value of: (i) a²+b² (ii) a²+ab+b²
(i) a²+b² = (2)²+(-2)² [Putting a = 2,b =-2 ] = 4 + 4 = 8 (ii) a²+ab+b² = (2)²+(2)(-2)+(-2)² [Putting a = 2,b =-2 ] = 4 – 4 + 4 = 4 Class 7 Maths Chapter 12 Exercise 12.3 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-12/
(i) a²+b² = (2)²+(-2)² [Putting a = 2,b =-2 ]
= 4 + 4 = 8
(ii) a²+ab+b²
= (2)²+(2)(-2)+(-2)² [Putting a = 2,b =-2 ]
= 4 – 4 + 4 = 4
Class 7 Maths Chapter 12 Exercise 12.3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-12/
Find the value of the following expressions, when x = -1: (i) 2x-7 (ii) -x + 2
(i) 2x-7 = 2(-1)-7 [Putting x =-1] = -2-7 = -9 (ii) -x + 2 = -(-1)+2 [Putting x =-1] = 1 + 2 = 3 Class 7 Maths Chapter 12 Exercise 12.3 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-12/
(i) 2x-7 = 2(-1)-7 [Putting x =-1]
= -2-7 = -9
(ii) -x + 2 = -(-1)+2 [Putting x =-1]
= 1 + 2 = 3
Class 7 Maths Chapter 12 Exercise 12.3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-12/
If p = -2, find the value of: (i) 4p+7 (ii) -3p²+4p+7
(i) 4p + 7 = 4(-2)+7 [Putting p =-2 ] = -8+7 = -1 (ii) -3p²+4p+7 = -3(-2)²+4(-2)+7 [Putting p =-2 ] = -3x4-8+7 = -12-8+7 = -20+7 = -13 Class 7 Maths Chapter 12 Exercise 12.3 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-12/
(i) 4p + 7 = 4(-2)+7 [Putting p =-2 ]
= -8+7 = -1
(ii) -3p²+4p+7
= -3(-2)²+4(-2)+7 [Putting p =-2 ]
= -3×4-8+7
= -12-8+7
= -20+7 = -13
Class 7 Maths Chapter 12 Exercise 12.3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-12/
If m = 2, find the value of: (i) m-2 (ii) 3m-5
(i) m - 2 = 2-2 [Putting m = 2 ] = 0 (ii) 3m-5 = 3x2-5 [Putting m = 2 ] = 6 – 5 = 1 Class 7 Maths Chapter 12 Exercise 12.3 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-12/
(i) m – 2 = 2-2 [Putting m = 2 ]
= 0
(ii) 3m-5 = 3×2-5 [Putting m = 2 ]
= 6 – 5 = 1
Class 7 Maths Chapter 12 Exercise 12.3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-12/
From the sum of 3x – y + 11 and – y – 11, subtract 3x – y – 11.
From the sum of (3x – y + 11) + (– y – 11) - (3x – y – 11) = 3x - y +11-y-11-3x+y+11 = 3x - 3x - y - y - y + 11 + 11 - 11 = (3-3)x - (1+111) y 111111 = 0x - y + 11 = -y+11 Class 7 Maths Chapter 12 Exercise 12.2 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/Read more
From the sum of
(3x – y + 11) + (– y – 11) – (3x – y – 11) = 3x – y +11-y-11-3x+y+11
= 3x – 3x – y – y – y + 11 + 11 – 11
= (3-3)x – (1+111) y 111111
= 0x – y + 11 = -y+11
Class 7 Maths Chapter 12 Exercise 12.2
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-12/