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`        Given,5a–3b / a  =  4a+b-2c / a+4b-2c = a+2b-3c / 4a-4c.Prove that 6a= 4b= 3c.`
2 months ago

```							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 get2a-2b-c = k(4a-4b-2c)=>2a-2b-c = 2k(2a-2b-c)=> k  = 1/2Thus, each ratio, k = 1/2.Now substituting value of , back in equation(1), we getb/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 getc/a = 2---(**)From, (*) and (**), we get 6a =4b = 3c
```
2 months ago
```							Dear student Let 6 a = 4 b = 3 c = 12k (12 is the LCM of 6,4 and 3)Hence,a = 2k b = 3k  c = 4k. { 5 a - 3 b }/ { a } =  { 5 ( 2 k ) - 3 ( 3 k ) } /{ 2 k } =  { k }/ { 2 k } =  { 1 }/ { 2 } { 4 a + b - 2 c }/ { a + 4 b - 2 c } =  { 4 ( 2 k ) + ( 3 k ) - 2 ( 4 k ) }/ { 2 k + 4 ( 3 k ) - 2 ( 4 k )}= {3k}/{6k}= {1}/{2} { a + 2 b - 3 c }/ { 4 a - 4 c } =  { 2 k + 2 ( 3 k ) - 3 ( 4 k ) }/ { 4 ( 2 k ) - 4 ( 4 k ) }= {-4k}/{-8k}= {1}/{2}Since all expressions are equal to {1}/{2}, therefore the assumption is correct.Therefore, 6a = 4b = 3c
```
2 months ago
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