Question

(5a-3b)/a=(4a+b-2c)/(a+4b-2c)=(a+2b-2c)/(4a-4c) then prove that 6a=4b=3c

Answer 1

Answer:

proved

Step-by-step explanation:

Given (5a -3b)/a = (4a+b-2c)/(a+4b-2c) = (a+2b-3c)/(4a-4c) then prove that 6a=4b=3c

Let us consider 6 a = 4 b = 3 c

We need to take LCM of 6, 4 and 3 , we get 12

So 6 a = 12 , a = 2

     4 b = 12, b = 3

      3 c = 12, c = 4

Substituting the given values we get

5 a - 3 b / a = 5(2) - 3(3) / 2 = 1 / 2

4 a + b - 2 c / a + 4 b - 2 c = 4(2) + 3 - 2(4) / 2 + 4(3) - 2(4) = 1 / 2

a + 2 b - 3 c / 4 a - 4 c = 2 + 2(3) - 3(4) = 1 /2

Thus all three are equal to 1 / 2.

Answer 2

Let Suppose that 6a=4b=3c=12x [Because 12 is the LCM of 6, 4, 3]

a=2x

b=3x

c=4x

Let’s start with Equation (1)

= (5a-3b)/a =({5(2x) - 3(3x)}/2x)

= (10x - 9x)/2x

= 1/2

Equation (2)

= (4a+b-2c)/ (a+ 4b-2c)

= {4(2x) +3x-2(4x)}/{2x+4(3x)-2(4x)}

= (3x)/(6x)

= 1/2

Equation (3)

= (a+2b-3c)/ (4a- 4c)

= {2x+2(3x)-3(4x)}/ {4(2x)-4(4x)}

= (-4x)/(-8x)

=1/2

Equation (1), (2) and (3) evaluates to the same value of 1/2.

So, our assumption 6a = 4b = 3c is Correct.

Hence Proved.