Question

A-4B+2C=16, 2A+7B-5C=83,find A+B-c=answer

Answer 1

Answer:

33

Explanation:(A-4B+2C) + (2A+7B-5C) = 16+83

3A+3B-3C= 99

3(A+B-C) = 99

A+B-C = 99÷3=33

Answer 2

Answer:

$A+B-C=\frac{148+2C}{5}+\frac{17+3C}{5} -C =(\frac{148+2C+17+3C-5C}{5} )=\frac{165}{5}=33$

SOLUTION:

  $A-4B+2C=16$ ………………. (i)

 $2A+7B-5C=83$…………… (ii)

Multiplying (i) by 2,

 $2A-8B+4C=32$

Now,  

$32+8B-4C=83-7B+5C$

Substituting value in (i) we get

 $A=16+4B-2C= 16+\frac{4(17+3C)}{5}-2C = \frac{80+68+12C-10C}{5} = \frac{148+2C}{5}$

Hence,

 $A+B-C=\frac{148+2C}{5}+\frac{17+3C}{5} -C =(\frac{148+2C+17+3C-5C}{5} )=\frac{165}{5}=33$