Question

3(a+3b)=11ab, 3(2a+b)=7ab solve by cross multiplication method​

Answer 1

Answer:

Given,

3 (a + 3b) = 11ab

→ 3a + 9b = 11ab (i)

3 (2a + b) = 7ab

→ 6a + 3b = 7ab (ii)

Now, dividing both sides of (i) and (ii) by ab, where both a and b are non-zero, we get

3/b + 9/a 11

→ 9/a + 3/b = 11 (iii)

6/b + 3/a = 7

→ 3/a + 6/b = 7 (iv)

Let us take, 1/a = x and 1/b = y. Then, (iii) and (iv) become

9x + 3y = 11

→ 3x + y = 11/3 (v)

3x+6y=7 (iv)

⇒5y = (21-11)/3

⇒ 5y = 10/3

⇒ y = 2/3

→ 1/b = 2/3 [: y = 1/b]

⇒ b = 3/2

Putting b = 3/2 in (iii), we get

9/a + 3/(3/2) = 11

⇒9/a + 2 = 11

⇒9/a 11 - 2

→ 9/a = 9

⇒ 1/a = 1

⇒a=1

:: the required solution be

a = 1 and b = 3/2

From (iv) and (v), on subtraction, we get

5y = 7 - 11/3