Question

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

Answer 1

this is right answer...

Answer 2

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

Solution:

$A-4B+2C=16$ ………………. (i)

$2A+7B-5C=83$ …………… (ii)

Multiplying (i) by 2,

$2A -8B+4C=32$

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

$\Rightarrow 8 \mathrm{B}-4 \mathrm{C}+7 \mathrm{B}-5 \mathrm{C}=83-32\Rightarrow \quad 15 \mathrm{B}-9 \mathrm{C}=51 \\\\\Rightarrow \quad 5 \mathrm{B}-3 \mathrm{C}=17 \\\\\Rightarrow \quad 5 \mathrm{B}=17+3 \mathrm{C} \\\\\Rightarrow \quad \mathrm{B}=\frac{17+3 C}{5}$

Substituting value in (i) we get

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

Hence,

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