Exercise 7.3
1. Place the last two terms of the following expressions in parentheses preceded by a minus sign:
(i) x + y – 3z + y
(ii) 3x – 2y – 5z – 4
(iii) 3a – 2b + 4c – 5
(iv) 7a + 3b + 2c + 4
(v) 2a2 – b2 – 3ab + 6
(vi) a2 + b2 – c2 + ab – 3ac
Solution:
we get,
(i) x + y – 3z + y = x + y – (3z – y)
(ii) 3x – 2y – 5z – 4 = 3x – 2y – (5z + 4)
(iii) 3a – 2b + 4c – 5 = 3a – 2b – (–4c + 5)
(iv) 7a + 3b + 2c + 4 = 7a + 3b – (–2c – 4)
(v) 2a2 – b2 – 3ab + 6 = 2a2 – b2 – (3ab – 6)
(vi) a2 + b2 – c2 + ab – 3ac = a2 + b2 – c2 – (- ab + 3ac)
2. Write each of the following statements by using appropriate grouping symbols:
(i) The sum of a – b and 3a – 2b + 5 is subtracted from 4a + 2b – 7.
(ii) Three times the sum of 2x + y – [5 – (x – 3y)] and 7x – 4y + 3 is subtracted from 3x – 4y + 7
(iii) The subtraction of x2 – y2 + 4xy from 2×2 + y2 – 3xy is added to 9×2 – 3y2- xy.
Solution:
(i) Given the sum of a – b and 3a – 2a = [(a-b) + (3a – 2b + 5)]
This is subtracted from 4a+ 2b – 7.
(ii) Given three times the sum of 2x + y {5- (x – 3y)} and 7x – 4y + 3 = 3 [(2x + y {5 –(x – 3y) } + (7x – 4y + 3)]
This is subtracted from 3x – 4y + 7.
(iii) Given the product of subtraction of x2 – y2 + 4xy from 2x2 + y2 – 3xy is given by {(2x3 + y2 – 3xy) – (x2 – y2 + 4xy)}
When the above equation is added is to 9x2 – 3y2 – xy we get
{(2x2 + y2 – 3xy) – (y2 – y2 + 4xy)} + (9x2 – 3y2– xy))
This is the required expression.