Question

Add 5a+2b-4c ,
and a-4b +3c
​

Answer 1

Step-by-step explanation:

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---(4)   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---(5)   From, (4) and (5), we get  6a =4b = 3c.Read more on Sarthaks.com - https://www.sarthaks.com/467526/given-5a-3b-a-4a-b-2c-a-4b-2c-a-2b-3c-4a-4c-prove-that-6a-4b-3c?show=467554#a467554

Answer 2

Answer:

$(5a + a) + ( - 4b + 2b) + ( - 4c + 2c) \\ = 4a - 2b - 2c \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

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