NCERT/CBSE MATHEMATICS CLASS 10 ... - TutorBreeze.com

NCERT/CBSE MATHEMATICS CLASS 10 textbook(OPTIONAL EXERCISE) http://www.TutorBreeze.com Contact for Online Tutoring in Physics, Math, Chemistry, English Answers to NCERT/CBSE MATHEMATICS Class 10(Class XI)textbook OPTIONAL EXERCISE CHAPTER THREE LINEAR EQUATIONS IN TWO VARIABLES EXERCISES

7. Solve the following pair of linear equations:

(i) px + qy = p – q

(ii) ax + by = c bx + ay = 1 + c

qx – py = p + q

(iv) (a – b)x + (a + b) y = a2 – 2ab – b2

x y − =0 . a b 2 2 ax + by = a + b

(iii )

(v) 152x – 378y = – 74 –378x + 152y = – 604

©TutorBreeze.com Please do not copy the answer given here Write to us for help in understanding the solution

(a + b)(x + y) = a2 + b2

NCERT/CBSE MATHEMATICS CLASS 10 textbook(OPTIONAL EXERCISE) http://www.TutorBreeze.com

Contact for Online Tutoring in Physics, Math, Chemistry, English

SOLUTION: (i) We shall solve the given system of equations by the method of cross- multiplication

(i) px + qy = p - q qx - py = p + q x y = −q ( p + q ) − p ( p − q ) -q(p-q)+p(p+q) x y ⇒ = 2 2 2 −qp − q − p + pq −qp + q + p 2 + pq x y 1 ⇒ 2 = 2 = 2 2 2 2 −q − p q +p -p -q x y 1 ⇒ = 2 = 2 2 2 2 −(q + p ) q + p − ( q + p2 ) ⇒

⇒

x y = − (1) 1

=

1 − (1)

⇒ x = 1; y = −1 Thus ,the required solution is x = 1andy = −1


=

1 -p -q 2 1 = 2 2 -p -q 2



SOLUTION: (ii) We shall solve the given system of equations by the method of cross- multiplication




(ii) ax + by = c bx + ay = 1 + c x y 1 = = 2 2 −b(1 + c) + ac −bc + a (1 + c) a − b x y 1 ⇒ = = 2 2 (−b − bc + ac) −bc + (a + ac) a − b x y 1 ⇒ = = 2 2 −b + c ( a − b ) a + c ( a − b ) a − b ⇒

⇒x=

−b + c ( a − b ) c b = − 2 2 a −b a + b (a − b)(a + b)

a + c ( a − b) c a = + 2 2 a −b a + b (a − b)(a + b) c b c a − ;y= + Thus ,the required solution is x = a + b (a − b)(a + b) a + b (a − b)(a + b) y=

SOLUTION: (iii)We shall solve the given system of equations by the method of cross- multiplication




x y − =0 a b ax + by = a 2 + b 2 x y 1 ⇒ = = 2 2 2 2 2 a ( a + b ) − 0 b( a + b ) + 0 a + b 2 x y 1 ⇒ = = 2 2 2 2 2 a ( a + b ) b( a + b ) a + b 2 a(a 2 + b 2 ) b( a 2 + b 2 ) ;y= 2 ⇒x= 2 a + b2 a + b2 ⇒ x = a; y = b Thus ,the required solution is x = a; y = b

SOLUTION: (iv) We shall solve the given system of equations by the method of cross- multiplication


NCERT/CBSE MATHEMATICS CLASS 10 textbook(OPTIONAL EXERCISE) http://www.TutorBreeze.com 2

(a - b)x + (a + b) y = a - 2ab - b

Contact for Online Tutoring in Physics, Math, Chemistry, English 2

. (a + b)(x + y) = a 2 + b 2 x ⇒ 2 2 −(a + b)(a + b ) + (a + b)(a 2 − 2ab − b 2 ) y = 2 2 (a − b)(a + b ) + (a + b)(a 2 − 2ab − b 2 ) 1 = (a − b)(a + b) − (a + b) 2 x ⇒ 2 2 (a + b)(− a − b + a 2 − 2ab − b 2 ) y = 3 2 2 2 − a + 2a b + ab − a b + 2ab3 − b3 + a 3 + ab 2 − a 2b − b3 1 = 2 2 a − b − a 2 − b 2 − 2ab x y 1 ⇒ = = 2 2 (a + b)(−2ab − 2b ) 4ab −2b(a + b) ⇒x=

(a + b)(−2ab − 2b 2 ) = a+b −2b(a + b)

4ab 2 2ab y= =− −2b(a + b) ( a + b) Thus the required solution is x = a + b and y = −

2ab ( a + b)

SOLUTION: (v) 152x – 378y = – 74………..(1) –378x + 152y = – 604……………..(2) Please note the peculiarity of the coefficients of x and y in these equations . The coefficients of x and y add up to the same number (-226) . Let us begin by adding the two equations, we get -226 x -226 y = - 678 ⇒ x + y = 3……….(3) Subtracting the two equations, we get -530x + 530 y = -530 ⇒ x – y = 1……………(4) Adding (3) and (4) , we get 2 x= 4 ⇒x = 2 ©TutorBreeze.com Please do not copy the answer given here Write to us for help in understanding the solution



Putting x back in (4) , we get y = 1. Thus the required solution is x = 2 and y = 1
