Question

maths 10th chapter 3 exercise 3.6 question 1​

Answer 1

Answer:

Solve the following pairs of equations by reducing them to a pair of linear equations:

(i) 1/2x + 1/3y = 2

(i) 1/2x + 1/3y = 21/3x + 1/2y = 13/6

Solution:

Let us assume 1/x = m and 1/y = n , then the equation will change as follows.

m/2 + n/3 = 2

m/2 + n/3 = 2⇒ 3m+2n-12 = 0…………………….(1)

m/2 + n/3 = 2⇒ 3m+2n-12 = 0…………………….(1)m/3 + n/2 = 13/6

m/2 + n/3 = 2⇒ 3m+2n-12 = 0…………………….(1)m/3 + n/2 = 13/6⇒ 2m+3n-13 = 0……………………….(2)

m/2 + n/3 = 2⇒ 3m+2n-12 = 0…………………….(1)m/3 + n/2 = 13/6⇒ 2m+3n-13 = 0……………………….(2)Now, using cross-multiplication method, we get,

m/2 + n/3 = 2⇒ 3m+2n-12 = 0…………………….(1)m/3 + n/2 = 13/6⇒ 2m+3n-13 = 0……………………….(2)Now, using cross-multiplication method, we get,m/(-26-(-36) ) = n/(-24-(-39)) = 1/(9-4)

m/2 + n/3 = 2⇒ 3m+2n-12 = 0…………………….(1)m/3 + n/2 = 13/6⇒ 2m+3n-13 = 0……………………….(2)Now, using cross-multiplication method, we get,m/(-26-(-36) ) = n/(-24-(-39)) = 1/(9-4)m/10 = n/15 = 1/5

m/2 + n/3 = 2⇒ 3m+2n-12 = 0…………………….(1)m/3 + n/2 = 13/6⇒ 2m+3n-13 = 0……………………….(2)Now, using cross-multiplication method, we get,m/(-26-(-36) ) = n/(-24-(-39)) = 1/(9-4)m/10 = n/15 = 1/5m/10 = 1/5 and n/15 = 1/5

m/2 + n/3 = 2⇒ 3m+2n-12 = 0…………………….(1)m/3 + n/2 = 13/6⇒ 2m+3n-13 = 0……………………….(2)Now, using cross-multiplication method, we get,m/(-26-(-36) ) = n/(-24-(-39)) = 1/(9-4)m/10 = n/15 = 1/5m/10 = 1/5 and n/15 = 1/5So, m = 2 and n = 3

m/2 + n/3 = 2⇒ 3m+2n-12 = 0…………………….(1)m/3 + n/2 = 13/6⇒ 2m+3n-13 = 0……………………….(2)Now, using cross-multiplication method, we get,m/(-26-(-36) ) = n/(-24-(-39)) = 1/(9-4)m/10 = n/15 = 1/5m/10 = 1/5 and n/15 = 1/5So, m = 2 and n = 31/x = 2 and 1/y = 3

m/2 + n/3 = 2⇒ 3m+2n-12 = 0…………………….(1)m/3 + n/2 = 13/6⇒ 2m+3n-13 = 0……………………….(2)Now, using cross-multiplication method, we get,m/(-26-(-36) ) = n/(-24-(-39)) = 1/(9-4)m/10 = n/15 = 1/5m/10 = 1/5 and n/15 = 1/5So, m = 2 and n = 31/x = 2 and 1/y = 3x = 1/2 and y = 1/3

m/2 + n/3 = 2⇒ 3m+2n-12 = 0…………………….(1)m/3 + n/2 = 13/6⇒ 2m+3n-13 = 0……………………….(2)Now, using cross-multiplication method, we get,m/(-26-(-36) ) = n/(-24-(-39)) = 1/(9-4)m/10 = n/15 = 1/5m/10 = 1/5 and n/15 = 1/5So, m = 2 and n = 31/x = 2 and 1/y = 3x = 1/2 and y = 1/3(ii) 2/√x + 3/√y = 2

m/2 + n/3 = 2⇒ 3m+2n-12 = 0…………………….(1)m/3 + n/2 = 13/6⇒ 2m+3n-13 = 0……………………….(2)Now, using cross-multiplication method, we get,m/(-26-(-36) ) = n/(-24-(-39)) = 1/(9-4)m/10 = n/15 = 1/5m/10 = 1/5 and n/15 = 1/5So, m = 2 and n = 31/x = 2 and 1/y = 3x = 1/2 and y = 1/3(ii) 2/√x + 3/√y = 24/√x + 9/√y = -1

m/2 + n/3 = 2⇒ 3m+2n-12 = 0…………………….(1)m/3 + n/2 = 13/6⇒ 2m+3n-13 = 0……………………….(2)Now, using cross-multiplication method, we get,m/(-26-(-36) ) = n/(-24-(-39)) = 1/(9-4)m/10 = n/15 = 1/5m/10 = 1/5 and n/15 = 1/5So, m = 2 and n = 31/x = 2 and 1/y = 3x = 1/2 and y = 1/3(ii) 2/√x + 3/√y = 24/√x + 9/√y = -1Solution: