Given: 2x² + x – a = 5 ⇒ 2(0)² + 0-a = 5 [Putting x = 0 ] ⇒ 0 + 0 - a = 5 ⇒ a = -5 Hence, the value of a is -5. Class 7 Maths Chapter 12 Exercise 12.3 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-12/
Given: 2x² + x – a = 5
⇒ 2(0)² + 0-a = 5 [Putting x = 0 ]
⇒ 0 + 0 – a = 5
⇒ a = -5
Hence, the value of a is -5.
(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
Let q should be subtracted. Then according to question, 3x²+4y²+5xy+20-q = -x² – y² +6xy+20? ⇒ q= 3x²+4y²+5xy+20-(-x² – y² +6xy+20) ⇒ q= 3x²+4y²+5xy+20 + x² + y² + 6xy-20 ⇒ q= 3x² +x²-4y²+y²+5xy-6xy+20-20 ⇒ q= 4x²-3y²-xy+0 Hence, 4x²-3y²-xy should be subtracted. Class 7 Maths Chapter 12 Exercise 12.Read more
Let q should be subtracted.
Then according to question,
3x²+4y²+5xy+20-q = -x² – y² +6xy+20?
⇒ q= 3x²+4y²+5xy+20-(-x² – y² +6xy+20)
⇒ q= 3x²+4y²+5xy+20 + x² + y² + 6xy-20
⇒ q= 3x² +x²-4y²+y²+5xy-6xy+20-20
⇒ q= 4x²-3y²-xy+0
Hence, 4x²-3y²-xy should be subtracted.
Simplify the expression and find its value when a= 5 and b =-3: 2(a²+ab)+3-ab
Given: 2(a²+ab)+3-ab ⇒ 2a²+2ab+3-ab ⇒ 2a²+2ab-ab+3 ⇒ 2a²+ab+3 ⇒ 2(5)² + (5)(-3)+3 [Putting a = 5 , b = -3 ] ⇒ 2 x 25 - 15 +3 ⇒ 50 - 15 + 3 ⇒ 38 Class 7 Maths Chapter 12 Exercise 12.3 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-12/
Given: 2(a²+ab)+3-ab
⇒ 2a²+2ab+3-ab
⇒ 2a²+2ab-ab+3
⇒ 2a²+ab+3
⇒ 2(5)² + (5)(-3)+3 [Putting a = 5 , b = -3 ]
⇒ 2 x 25 – 15 +3
⇒ 50 – 15 + 3
⇒ 38
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/
What should be the value of a if the value of 2x² + x – a equals to 5, when x = 0 ?
Given: 2x² + x – a = 5 ⇒ 2(0)² + 0-a = 5 [Putting x = 0 ] ⇒ 0 + 0 - a = 5 ⇒ a = -5 Hence, the value of a is -5. Class 7 Maths Chapter 12 Exercise 12.3 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-12/
Given: 2x² + x – a = 5
⇒ 2(0)² + 0-a = 5 [Putting x = 0 ]
⇒ 0 + 0 – a = 5
⇒ a = -5
Hence, the value of a is -5.
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/
Simplify the expressions and find the value if x is equal to 2: x+7+4(x-5)
x+7+4(x-5) = x+7+4x-20 = x + 4x + 7 - 20 = 5x - 13 = 5 x 2 - 13 [Putting x = 2 ] = 10 - 13 = -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/
x+7+4(x-5) = x+7+4x-20 = x + 4x + 7 – 20
= 5x – 13 = 5 x 2 – 13 [Putting x = 2 ]
= 10 – 13 = -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/
When a = 0,b=-1, find the value of the given expressions: 2a+2b
(i) 2a + 2b = 2(0)+2(-1) [Putting a = 0,b = -1] = 0 – 2 = -2 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) 2a + 2b = 2(0)+2(-1) [Putting a = 0,b = -1]
= 0 – 2 = -2
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 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/
What should be taken away from 3x² +4y²+5xy+20 to obtain -x² – y² +6xy+20?
Let q should be subtracted. Then according to question, 3x²+4y²+5xy+20-q = -x² – y² +6xy+20? ⇒ q= 3x²+4y²+5xy+20-(-x² – y² +6xy+20) ⇒ q= 3x²+4y²+5xy+20 + x² + y² + 6xy-20 ⇒ q= 3x² +x²-4y²+y²+5xy-6xy+20-20 ⇒ q= 4x²-3y²-xy+0 Hence, 4x²-3y²-xy should be subtracted. Class 7 Maths Chapter 12 Exercise 12.Read more
Let q should be subtracted.
Then according to question,
3x²+4y²+5xy+20-q = -x² – y² +6xy+20?
⇒ q= 3x²+4y²+5xy+20-(-x² – y² +6xy+20)
⇒ q= 3x²+4y²+5xy+20 + x² + y² + 6xy-20
⇒ q= 3x² +x²-4y²+y²+5xy-6xy+20-20
⇒ q= 4x²-3y²-xy+0
Hence, 4x²-3y²-xy should be subtracted.
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/