Chapter 12 Algebraic Expressions Exercise 12.3

  Question 2.
  
  If p = -2, find the value of:
  
  (i) 4p + 7
  
  (ii) -3p2 + 4p + 7
  
  (iii) -2p3 – 3p2 + 4p + 7
  
  Solution:
  
  (i) 4p + 7
  
  Putting p = -2, we get 4(-2) + 7 = -8 + 7 = -1
  (ii) -3p2 + 4p + l
  
  Putting p = -2, we get
  
  -3(-2)2 + 4(-2) + 7
  
  = -3 × 4 – 8 + 7 = -12 – 8+ 7 = -13
  (iii) -2p3 – 3p2 + 4p + 7
  
  Putting p = -2, we get
  
  – 2(-2)3 – 3(-2)2 + 4(-2) + 7
  
  = -2 × (-8) – 3 × 4 – 8 + 7
  
  = 16 – 12 – 8 + 7 = 3
  Question 3.
  
  If a = 2, b = -2, find the value of:
  
  (i) a2 + b2
  
  (ii) a2 + ab + b2
  
  (iii) a2 – b2
  
  Solution:
  
  (i) a2 + b2
  
  Putting a = 2 and b = -2, we get
  
  (2)2 + (-2)2 = 4 + 4 = 8
  (ii) a2 + ab + b2
  
  Putting a = 2 and b = -2, we get
  
  (2)2 + 2(-2) + (-2)2 = 4 – 4 + 4 = 4
  (iii) a2 – b2
  
  Putting a = 2 and b = -2, we get
  
  (2)2 – (-2)2 = 4 – 4 = 0
  Question 4.
  
  When a = 0, b = -1, find the value of the given expressions:
  
  (i) 2a + 2b
  
  (ii) 2a2 + b2 + 1
  
  (iii) 2a2b + 2ab2 + ab
  
  (iv) a2 + ab + 2
  
  Solution:
  
  (i) 2a + 2b = 2(0) + 2(-1)
  
  = 0 – 2 = -2 which is required.
  (ii) 2a2 + b2 + 1
  
  = 2(0)2 + (-1)2 + 1 =0 + 1 + 1 = 2 which is required.
  (iii) 2a2b + 2ab2 + ab
  
  = 2(0)2 (-1) + 2(0)(-1)2 + (0)(-1)
  
  =0 + 0 + 0 = 0 which is required.
  (iv) a2 + ab + 2
  
  = (0)2 + (0)(-1) + 2
  
  = 0 + 0 + 2 = 0 which is required.
  Question 5.
  
  Simplify the expressions and find the value if x is equal to 2.
  
  (i) x + 7 +4(x – 5)
  
  (ii) 3(x + 2) + 5x – 7
  
  (iii) 6x + 5(x – 2)
  
  (iv) 4(2x – 1) + 3x + 11
  
  Solution:
  
  (i) x + 7 + 4(x – 5) = x + 7 + 4x – 20 = 5x – 13
  
  Putting x = 2, we get
  
  = 5 × 2 – 13 = 10 – 13 = -3
  
  which is required.
  (ii) 3(x + 2) + 5x – 7 = 3x + 6 + 5x -7 = 8x – 1
  
  Putting x = 2, we get
  
  = 8 × 2 – 1 = 16 – 1 = 15
  
  which is required.
  (iii) 6x + 5(x – 2) = 6x + 5x – 10
  
  = 11 × – 10
  
  Putting x = 2, we get
  
  = 11 × 2 – 10 = 22 – 10 = 12
  
  which is required.
  (iv) 4(2x – 1) + 3x + 11 = 8x – 4 + 3x + 11
  
  = 11x + 7
  
  Putting x = 2, we get
  
  = 11 × 2 + 7 = 22+ 7 = 29
  Question 6.
  
  Simplify these expressions and find their values if x = 3, a = -1, b = -2.
  
  (i) 3x – 5 – x + 9
  
  (ii) 2 – 8x + 4x + 4
  
  (iii) 3a + 5 – 8a + 1
  
  (iv) 10 – 3b – 4 – 55
  
  (v) 2a – 2b – 4 – 5 + a
  
  Solution:
  
  (i) 3x – 5 – x + 9 = 2x + 4
  
  Putting x = 3, we get
  
  2 × 3 + 4 = 6 + 4 = 10
  
  which is required.
  (ii) 2 – 8x + 4x + 4 = -8x + 4x + 2 + 4 = -4x + 6
  
  Putting x = 2, we have
  
  = -4 × 2 + 6 = -8 + 6 =-2
  
  which is required.
  (iii) 3a + 5 – 8a +1 = 3a – 8a + 5 + 1 = -5a + 6
  
  Putting a = -1, we get
  
  = -5(-1) + 6 = 5 + 6 = 11
  
  which is required.
  (iv) 10 – 3b – 4 – 5b = -3b – 5b + 10 – 4
  
  = -8b + 6
  
  Putting b = -2, we get
  
  = -8(-2) + 6 = 16 + 6 = 22
  
  which is required.
  (v) 2a – 2b – 4 – 5 + a = 2a + a – 2b – 4 – 5
  
  = 3a – 26 – 9
  
  Putting a = -1 and b = -2, we get
  
  = 3(-1) – 2(-2) – 9
  
  = -3 + 4 – 9 = 1 – 9 = -8
  
  which is required.
  Question 7.   
   (i) If z = 10, find the value of z2 – 3(z – 10).   
   (ii) If p = -10, find the value of p2 -2p – 100.   
   Solution:   
   (i) z2 – 3(z – 10)   
   = z2 – 3z + 30   
   Putting z = 10, we get   
   = (10)2 – 3(10) + 30   
   = 1000 – 30 + 30 = 1000 which is required. 
  (ii) p2 – 2p – 100
  
  Putting p = -10, we get
  
  (-10)2 – 2(-10) – 100
  
  = 100 + 20 – 100 = 20 which is required.
  Question 8.
  
  What should be the value of a if the value of 2x2 + x – a equals to 5, when x = 0?
  
  Solution:
  
  2x2 + x – a = 5
  
  Putting x = 0, we get
  
  2(0)2 + (0) – a = 5
  
  0 + 0 – a = 5
  
  -a = 5
  
  ⇒ a = -5 which is required value.
  Question 9.
  
  Simplify the expression and find its value when a = 5 and b = -3.
  
  2(a2 + ab) + 3 – ab
  
  Solution:
  
  2(a2 + ab) + 3 – ab = 2a2 + 2ab + 3 – ab
  
  = 2a2 + 2ab – ab + 3
  
  = 2ab + ab + 3
  
  Putting, a = 5 and b = -3, we get
  
  = 2(5)2 + (5)(-3) + 3
  
  = 2 × 25 – 15 + 3
  
  = 50 – 15 + 3
  
  = 53 – 15 = 38
  
  Hence, the required value = 38.



