# Ssamacheer Kalvi Math Solution for Exercise 3.11 Part2

This page samacheer kalvi math solution for exercise 3.11 part2 is going to provide you solution for every problems that you find in the exercise no 5.5

## Samacheer Kalvi Math Solution for Exercise 3.11 part2

(v) [(2x²-5x+3)/(x²-3x+2)]-[(2x²-7x-4)/(2x² - 3 x - 2)]

Solution:

= [(2x²-5x+3)/(x²-3x+2)]-[(2x²-7x-4)/(2x² - 3 x - 2)]

(vi) [(x²-4)/(x²+6x+8)]-[(x²-11x+30)/(x²-x - 20)]

Solution:

= [(x²-4)/(x²+6x+8)]-[(x²-11x+30)/(x²-x - 20)]

(x²-2²) = (x + 2) (x - 2)

(x²+ 6 x + 8) = (x + 2) (x + 4)

(x²-11x+30) = (x - 6) (x - 5)

(x²-x - 20) = (x - 5) (x + 4)

(vii) [(2x + 5)/(x + 1)] + [(x² + 1)/(x² - 1)] - [(3x - 2)/(x - 1)]

Solution:

(viii) [1/(x²+3x+2)] + [1/(x²+5x+6)] - [2/(x²+4x+3)]

Solution:

(x²+3x+2) = (x + 1) (x + 2)

(x²+5x+6) = (x + 2)(x + 3)

(x²+4x+3) = (x + 3) (x + 1)

= 0

In the page samacheer kalvi math solution for exercise 3.11 part1 we are going to see the solution of next problem

(2) Which rational expression should be added to (x³-1)/(x²+2) to get (3x³ + 2x² + 4)/(x² + 2)?

Solution:

let the required rational expression be p (x)

[(x³-1)/(x²+2)]  + p (x) = (3x³ + 2x² + 4)/(x² + 2)

p (x) = [(3x³ + 2x² + 4)/(x² + 2)] - [(x³-1)/(x²+2)]

= [(3x³ + 2x² + 4) - (x³-1)]/(x² + 2)

=  (3x³ - x³ + 2x² + 4 + 1)/(x² + 2)

=  (2 x³ + 2 x² + 5)/(x² + 2)

(3) Which rational expression should be subtracted from (4x³ - 7 x² + 5)/(2x - 1) to get 2x² - 5x + 1?

Solution:

let p(x) be the required rational expression

[(4x³ - 7 x² + 5)/(2x - 1)] - p(x) = 2x² - 5x + 1

[(4x³ - 7 x² + 5)/(2x - 1)] - (2x² - 5x + 1) = p(x)

p(x) = [(4x³ - 7 x² + 5) - (2x² - 5x + 1)(2x - 1)]/(2x - 1)

(4) If P = x/(x + y) Q = y/(x + y),then find [1/(P-Q)] - [2Q/(P²-Q²)]