Question

SOLVE 3(a + 3b) = 11ab , 3(2a +b) = 7ab

Answer 1

The answer is given below :

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)

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

5y = 7 - 11/3

⇒ 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

Hope it helps you.

Answer 2

Hello User !!!

Your Question is :

3 × ( a + 3b ) = 11ab   ......... Equation No. 1

3 × ( 2a + b ) = 7ab     ......... Equation No. 2

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Equation No. 1

3 × ( a + 3b ) = 11ab 

Opening Brackets

3a + 9b = 11ab

Divide both sides by "ab"

3a/ab + 9b/ab = 11ab/ab

3/b + 9/a = 11


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Equation No. 2

3 × ( 2a + b ) = 7ab

Opening Brackets 

6a + 3b = 7ab

Divide both sides by "ab"

6a/ab + 3b/ab = 7ab/ab

6/b + 3/a = 7

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Let 1/b be 'x' and 1/a be 'y'

So Equation No. 1 will be :

3x + 9y = 11

And Equation No. 2 will be :

6x + 3y = 7

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Use Elimination Method :

Let make Co efficient of 'y' to 9 in both New Equations

 Equation No. 1 :

3x + 9y = 11 It is already 9

Equation No. 2 :

6x + 3y = 7               ....× 3

18x + 9y = 21 Now, it is 9


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Using Elimination Method :

       3x + 9y = 11

(-)  18x + 9y = 21

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      -15x - 0 = -10

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15x = 10

x = 10/15 = 2/3


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Substituting 'x' in Equation No. 1 to get 'y'

3x + 9y = 11

3 × 2/3 + 9y = 11

2 + 9y = 11

9y = 9

y = 1


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Now Find 'a' and 'b'  :

x = 1/b

So,

2/3 = 1/b                                      [  We know that x = 2/3 ]

b = 3/2

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y = 1/a

So,

1 = 1/a                                            [ We know that y = 1 ]

a = 1


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So answer is :

a = 1 

b = 3/2 = 1.5