Friday, July 19, 2013

rationalisation


                                                                   



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Wednesday, July 17, 2013

proportion

Find the ratio of x:y:z from the following equations:

ax + by + cz = 0
lx + my + nz = 0



Answer

divide the 2 equations by z, we get,

a(x/z) + b(y/z) + c = 0

l(x/z) + m(y/z)+ n = 0

write as follows:

b                c               a             b

m               n                l             m   

cross multiply:
x/z = (bn - mc)/ (am - lb)


y/z = (cl - na) / (am - bl)


So, x/ (bn - mc) = y/ (cl - na) = z/ (am - lb)  answer


proportion

if a/b = c/d = e/f
then show that these ratios are equal to 
(5a - 7c + 3e) / (5b -7d + 3f)



Answer:

let  a/b = c/d = e/f = k
therefore, a = bk, c = dk and e = fk
substituting in  (5a - 7c + 3e)
we get, 5bk - 7dk + 3fk  = k(5b - 7d + 3f)

Therefore,
(5a - 7c + 3e) / (5b -7d + 3f) = k(5b -7d + 3f) / (5b -7d + 3f) = k
Hence (5a - 7c + 3e) / (5b -7d + 3f)  has same value as the ratios.







If p/q = r/s = t/u, then prove that (p2 – pr +t2)/(q2 – qs + u2) = pt/qu





Answer
let,  p/q = r/s = t/u = k
so, p = qk, r = sk and t = uk
therefore, (p2 – pr +t2) = q2k2 - qsk2 - u2k2 = k2(q2 - qs - u2)
therefore, (p2 – pr +t2)/(q2 – qs + u2) = k2(q2 - qs - u2)/ (q2 - qs - u2) = k2
= p/q x t/u
= pt/qu