Question

class 10 maths exercise 3.7 question number 7

Answer 1

Answer:

I DON'T KNOW

Step-by-step explanation:

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Answer 2

Answer:

Question- 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

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