Question

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

Answer 1

Let $6 a = 4 b = 3 c = 12k$ (12 is the LCM of 6,4 and 3)

Hence,

$a = 2k\\b = 3k\\c = 4k$.

1. $\frac { 5 a - 3 b } { a } = \frac { 5 ( 2 k ) - 3 ( 3 k ) } { 2 k } = \frac { k } { 2 k } = \frac { 1 } { 2 }$
2. $\frac { 4 a + b - 2 c } { a + 4 b - 2 c } = \frac { 4 ( 2 k ) + ( 3 k ) - 2 ( 4 k ) } { 2 k + 4 ( 3 k ) - 2 ( 4 k )}=\frac{3k}{6k}=\frac{1}{2}$
3. $\frac { a + 2 b - 3 c } { 4 a - 4 c } = \frac { 2 k + 2 ( 3 k ) - 3 ( 4 k ) } { 4 ( 2 k ) - 4 ( 4 k ) }=\frac{-4k}{-8k}=\frac{1}{2}$

Since all expressions are equal to $\frac{1}{2}$, therefore the assumption is correct.

Therefore, 6a = 4b = 3c

Answer 2

Answer:

Multiply the numerator and denominator of second fraction by -1and then do addendo.

Step-by-step explanation:

Hi,

Given (5a-3b) /a=(4a+b-2c) /(a+4b-2c) =(a+2b-3c) /(4a-4c)

Let each constant be equal to 'k'

=> (5a-3b) /a = k

=>  (5a-3b) = ka-------(1)

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

=>(4a+b-2c) = k(a+4b-2c)----(2)

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

=>(a+2b-3c) =k(4a-4c)-------(3)

Now, Adding (1)-(2)+(3), we get

2a-2b-c = k(4a-4b-2c)

=>2a-2b-c = 2k(2a-2b-c)

=> k  = 1/2

Thus, each ratio, k = 1/2.

Now substituting value of , back in equation(1), we get

b/a = 3/2---(*)

Substituting value of b/a = 3/2 and value of k =1/2 in any one of equations

(2) or (3), we get

c/a = 2---(**)

From, (*) and (**), we get 6a =4b = 3c.

Hope, it helped !