Question

If $41x+31y=18$ and $31x+47y=60$ then find the value of $x+y$.

• A. 0.9
• B. 1
• C. 0.2
• D. 0.8

Answer 1

The correct option is A 0.9

$(41x+31y=18)\ast 47\Rightarrow 41\ast 47x+31\ast 47y=18\ast \mathrm{47.............}\left(1\right)$

$(31x+47y=60)\ast 31\Rightarrow 31\ast 31x+31\ast 47y=60\ast \mathrm{31..............}\left(2\right)$

Now, Subtracting (2) from (1)

$41\ast 47x-31\ast 31x=18\ast 47-60\ast 31\Rightarrow 966x=846-1860=-1014\Rightarrow x=\frac{-1014}{966}=-1.05$

Putting this value in Original Equation

$41\ast (-1.05)+31\ast y=18\Rightarrow -43.05+31y=18\Rightarrow 31y=18+43.05=61.05\Rightarrow y=\frac{61.05}{31}=1.95$

So, $x+y=1.95-1.05=0.9$