0

Guest

Courses New

7. Solve the following pair of linear equations: (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 = a2 + b2 (iv) (a – b)x + (a + b) y = a2 – 2ab – b2 (a + b)(x + y) = a2 + b2 (v) 152x – 378y = – 74 –378x + 152y = – 604

7. Solve the following pair of linear equations:

(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 = a2 + b2

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

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

(v) 152x – 378y = – 74

–378x + 152y = – 604

Grade:12th pass

1 Answers

Pawan Prajapati
askIITians Faculty 60787 Points
2 years ago
Solutions: (i) px + qy = p – q……………(i) qx – py = p + q……………….(ii) Multiplying p to equation (1) and q to equation (2), we get p2x + pqy = p2 − pq ………… (iii) q2x − pqy = pq + q2 ………… (iv) Adding equation (iii) and equation (iv),we get p2x + q2 x = p2 + q2 (p2 + q2 ) x = p2 + q2 x = (p2 + q2)/ p2 + q2 = 1 From equation (i), we get p(1) + qy = p – q qy = p-q-p qy = -q y = -1 (ii) ax + by= c…………………(i) bx + ay = 1+ c………… ..(ii) Multiplying a to equation (i) and b to equation (ii), we obtain a2x + aby = ac ………………… (iii) b2x + aby = b + bc…………… (iv) Subtracting equation (iv) from equation (iii), (a2 – b2) x = ac − bc– b x = (ac − bc– b)/ (a2 – b2) x = c(a-b) –b / (a2+b2) From equation (i), we obtain ax +by = c a{c(a−b)−b)/ (a2 – b2)} +by=c ac(a−b)−ab/ (a2 – b2)+by=c by=c–ac(a−b)−ab/(a2 – b2) by=abc – b2 c+ab/a2-b2 y = c(a-b)+a/a2-b2 (iii) x/a – y/b = 0 ax + by = a2 + b2 x/a – y/b = 0 => bx − ay = 0 ……. (i) ax + by = a2 + b2 …….. (ii) Multiplying a and b to equation (i) and (ii) respectively, we get b2x − aby = 0 …………… (iii) a2x + aby = a 3 + ab3 …… (iv) Adding equations (iii) and (iv), we get b2x + a2x = a 3 + ab2 x (b2 + a2) = a (a2 + b2) x = a Using equation (i), we get b(a) − ay = 0 ab − ay = 0 ay = ab, y = b (iv) (a – b)x + (a + b) y = a2 – 2ab – b2 (a + b)(x + y) = a2 + b2 (a + b) y + (a – b) x = a2− 2ab − b2 …………… (i) (x + y)(a + b) = a 2 + b2 (a + b) y + (a + b) x = a2 + b2 ………………… (ii) Subtracting equation (ii) from equation (i), we get (a − b) x − (a + b) x = (a 2 − 2ab − b 2) − (a2 + b2) x(a − b − a − b) = − 2ab − 2b2 − 2bx = − 2b (b + a) x = b + a Substituting this value in equation (i), we get (a + b)(a − b) +y (a + b) = a2− 2ab – b2 a2 − b2 + y(a + b) = a2− 2ab – b2 (a + b) y = − 2ab y = -2ab/(a+b) (v) 152x − 378y = − 74 76x − 189y = − 37 x =(189y-137)/76……………..…(i) − 378x + 152y = − 604 − 189x + 76y = − 302 ………….. (ii) Using the value of x in equation (ii), we get −189(189y−37/76)+76y=−302 − (189)2y + 189 × 37 + (76)2 y = − 302 × 76 189 × 37 + 302 × 76 = (189)2 y − (76)2y 6993 + 22952 = (189 − 76) (189 + 76) y 29945 = (113) (265) y y = 1 Using equation (i), we get x = (189-37)/76 x = 152/76 = 2

Think You Can Provide A Better Answer ?

Other Related Questions on 10 grade maths

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More