Linear Equation in 2 variables

No need to solve the equations; first equation is the ans :- x + y = 3

1st equation is the solution, no need to solve it.

No need to solve... Que itself reveals the answer x+y=3

no need to solve the equaation as the answer lies in question itself!!

4(x+y=3) ==> 4x+4y=12 4x+4y=2x+5y 2x=y x+y = x+2x=3 ==> x=1

The reason why the second equation is necessary is because there are an infinite number of solutions to the first equation, A few examples: 1.1+1.9=3 1.01+1.99=3 0.25+1.75=3 Only (x=1,y=2) satisfies both equations

HAHAHA!!!!! This is a possible troll question!!! Don't need to find individual values! You don't even need the 2nd equation!! HAHAHA!!!! Troller!!!

Sum=first equation= $3$ .Common sense.

no nead to solve the equations x+y is already given

No need to use your stuff. Just use your common sense and give the answer. :D

Given away like that.

Funniest problem I solved today :D Answer is within the question!

Already given

$Using\quad elimination\quad method\quad we\quad can\quad solve\quad this\quad pair\quad of\quad linear\quad equation\\ x+y=3\quad \quad ....(i)\\ 2x+5y=12\quad ....(ii)\\ \quad \\ \quad \quad \quad \quad Multiplying\quad (i)\quad by\quad 2\quad to\quad eliminate\quad x\\ \quad \quad \quad \quad and\quad then\quad subtracting\quad (i)\quad from\quad (ii)\\ \\ \quad \quad \quad 2x\quad +\quad 5y\quad =\quad 12\\ \quad \quad \quad 2x\quad +\quad 2y\quad =\quad 6\\ \quad \quad \ominus \quad \quad \ominus \quad \quad \quad \quad \quad \ominus \\ \quad =\quad 0\quad \quad +\quad 3y\quad =\quad 6\\ \quad \quad \quad y=\frac { 6 }{ 3 } =2\\ \\ By\quad substituting\quad value\quad of\quad y\quad in\quad (i)\quad we\quad get\quad the\quad value\quad of\quad x\\ \quad \quad \quad \quad x\quad +(2)\quad =\quad 3\\ \quad \quad \quad \quad x\quad =\quad 1\\$