Question

5y2+40y-12/5y+10y2-4=y+8/1+2y​

Answer 1

Given :

(5y² + 40y - 12) / (5y + 10y² - 4) = (y + 8) / (1 + 2y)

To find :

Value of y

Solution :

(5y² + 40y - 12) / (5y + 10y² - 4) = (y + 8) / (1 + 2y)

(5y² + 40y - 12)(1 + 2y) = (y + 8)(5y + 10y² - 4)

5y² + 10y³ + 40y + 80y² - 12 - 24y = 5y² + 10y³ - 4y + 40y + 80y² - 32

- 12 - 24y = - 4y - 32

24y - 4y = 32 - 12

20y = 20

y = 1

Verification :

LHS :

(5y² + 40y - 12) / (5y + 10y² - 4)

(5 + 40 -12) / (5 +10 - 4)

33 / 11

3

RHS :

(y + 8) / (1 + 2y)

(1 + 8) / (1 + 2)

9 / 3

3

LHS = RHS

Hence y = 1

Answer 2

if a(y+z) = b(z+x) = c(x+y) and out of a,b,c on two of them are equal then show that, y-z = z-x = x-y/a (b-c) b(c-a) c(a-b)

Answer a(y+z) = b (z+x) = c(x+y)

y-z = z+x = x-y

a(b-c) b(c-a) c(a-b)

a (y+z) = b(z+x) = c(x+y)

abc abc abc

y+z = z+x = x+y = k

bc ac ab

x+y = k

ab-ac

similarly x-y = k

c(a-b)

x+y-(z-x) = k

a(b-c)

y-z

a(b-c)

y+z-(x+y)

bc-ab

zx = k-10

b(c-a)

y-z = z-x = x-y for 1,2 fo

a(b-c) b(-a) c(a-b)