Solve : px+qy = p ?qqx ?py = p+ q . Easy.
5/6/2017 · px + qy=p-q —–(1) qx- py=p + q —–(2) Multiplying (1) with p and (2) with q we get, p²x+pqy=p²-pq —–(3) q²x-pqy=pq+q² —–(4) Adding (3) with (4) we get, p²x+q²x=p²+q². or, x(p²+q²)=p²+q². or, x=1. Putting in (1) we get, p.1+ qy=p-q . or, qy=p-q -p. or, qy=- q.
Qx- Py=P + Q …..(2) Now multiplying equation (1) with P and equation (2) with Q we get. Px + Qy=P-Q ×P. Qx- Py=P + Q × Q =P× Px +PQy=P×P-QP…..(3) Q ×Qx-PQy=P× Q + Q × Q ….(4) Now by adding equation (3) and (4) we get (Psquare+Qsquare )x=Psquare+Qsquare. Implies x= Psquare+Qsquare /Psquar+Qsquare. Which implies x=1. Consider PX + Qy=P-Q =P(1)+ Q (y)=P- Q =P+ Qy=P-Q = Qy=P-Q -P, 8/18/2019 · Px + qy = p-q /qx- py=p + q , find and by substitution. ? is done on EduRev Study Group by Class 10 Students. The Questions and Answers of Give equation. Px + qy = p-q /qx- py=p + q , find and by substitution. ? are solved by group of students and teacher of Class 10, which is also the largest student community of Class 10.
5/12/2007 · Favorite Answer. multiply first equation by q . multiply second equation by p. so as to have a pqx term in both to be able to eliminate x. pqx + q ^2y = pq – q ^2 (equation 3) pqx – p^2y = p^2 + pq (equation 4) Subtract equation 4 from equation 3: q ^2y + p^2y = – q ^2 – p^2.
Saksham Verma, added an answer, on 8/9/12. Saksham Verma answered this. ( px + qy = p-q ) x ( q ) ( qx – py = p + q ) x (p) pqx + q 2 y = pq – q 2 – pqx -p 2 y = p 2 + pq = 0 + ( q 2 +p 2 )y = pq – q 2 – p 2 – pq. = ( q 2 + p 2 )y = – ( q 2 + p 2) y = -1. putting y =2 in 1. px + qy = p-q . px + q (-1) = p- q . px – q = p- q .
cocept of elimation method The sum of the digits of a two digit number is 12 . The number obtained by interchanging Its digits exceed the given number by 18 find the number x+2y=5 3x/2+3y=10 solve by elimination method Parthiv tells his son ArivFive years ago my age was seven times your age then.
(i) px + qy = p – q . qx – py = p + q (ii) ax + by = c. bx + ay = 1 + c (iii) x/a – y/b = 0. ax + by = a 2 + b 2 (iv) (a – b)x + (a + b) y = a 2 – 2ab – b 2 (a + b)(x + y) = a 2 + b 2 (v) 152x – 378y = – 74 –378x + 152y = – 604. Solutions: (i) px + qy = p – q ……………(i) qx – py = p + q ……………….(ii), 12/18/2020 · Ex 3.7, 7 Solve the following pair of linear equations: px + qy = p – q qx + py = p + q px + qy = p ? q Multiplying both sides by q q ( px + qy) = q (p – q ) pqx + q2y = pq ? q2 qx ? py = p + q Multiplying both sides by p p(qx + py) = p (p + q ) pqx + p2y = p2 + pq Hence, our equa, (i) px + qy = p – q , qx – py = p + q (ii) ax + by = c, bx + ay = 1 + c (iii) ax + by = a 2 + b 2 (iv) (a – b)x + (a + b)y = a 2 – 2ab – b 2, (a + b) (x + y) = a 2 + b 2 (v) 152x – 378y = -74, – 378x + 152y = -604 SOLUTION: Soln. : (i) We have : px + qy = p – q ….. (1) qx – py = p + q ….. (2) Multiply (1) by pand (2) by q , we get p 2 + qpy = p 2 – pq…. (3) q 2 – qpy = q 2 + pq…. (4)
