The solution of the linear equation should satisfy the equation. The given values of x and y are 7 and 3 respectively.

Among the given equations, only the equation x + 3y = 16 is able to satisfy the equation.
$\Rightarrow 7+3\left(3\right)=16$
Hence $x=7andy=3$ is the solution of   x + 3y = 16. 

If the point (2,2) lies on the given linear equation, then it must satisfy the equation.
Putting the value of x and y in the equation we get:
$\Rightarrow 4\left(2\right)+5\left(2\right)=k$
$\Rightarrow k=18$
Thus, the value of k is 18.

The solutions of a linear equation must be able to satisfy the equation. Among the given options, point (2,-1) satisfies the given equation. Substituing the value of x and y in the equation we get:
3(2) + 4(-1) = 2. 

The solution of a linear equation satisfies the equation.
 $3\times 3+4\times 4=9+16=25\ne 38\to$ (3, 4) does not satisfy the equation.
$3\times 6+4\times 5=18+20=38\to$ (6, 5) satisfies the equation.
$3\times 2+4\times 19=6+76=82\ne 38\to$ (2, 19) does not satisfy the equation.
$3\times 3+4\times 12=9+48=57\ne 38\to$ (3, 12) does not satisfy the equation.

Let $l$ and $b$ be the length and breadth of the room respectively.
Then, the perimeter of the room is $2(l+b)$ metres.
From the question, 
$l=6+b$    ...(1)
$\frac{1}{2}\times 2(l+b)=46$
$\Rightarrow l+b=46$    ...(2)
Let's solve these two equations using substitution method.
On substituting the value of $l$ from (1) in (2), we get
$6+b+b=46$
$\Rightarrow$ $6+2b=46$
$\Rightarrow 2b=40$
$\Rightarrow b=20m$
$\Rightarrow l=20+6=26m$
Therefore, length and breadth are 26 m and 20 m long respectively.

If the graph of linear equations represented by the lines intersect at only one point, then the point is its only solution.
Here, the lines meet at only one point (1,-1). Therefore, the given pair of equations has a unique solution.

The given equation is $y=\frac{1}{2}(3x+7)$.
Simplifying the equation we get
$2y-3x-7=0$ 
Rewriting the equation in standard form ax + by + c = 0 we get
$-3x+2y-7=0$
$\therefore$ On comparing the above equation with standard form ax + by +c = 0, we get the values of a, b and c is -3, 2 and -7 respectively.
But the equation can also be written as,
$3x-2y+7=0$
(By multiplying both the sides by (-1))
$\therefore$ On comparing the above equation with standard form ax + by +c = 0, we get the value of a, b and c is +3, -2 and +7 respectively.

Given
$2x-3y=7$...(i)
$5x+y=9$...(ii)
Rearranging (ii), we get  $y=9-5x$...(iii)
Substituting (iii) in (i),  we get
$2x-3(9-5x)=7$
$\Rightarrow$$17x=34$
$\Rightarrow x=2$.
Substituting the value of  $x$ in (i), we get
$2\left(2\right)-3y=7$
$\Rightarrow y=-1$.

$x+2y=5$   ...(1)

$7x+3y=13$  ...(2) 

Let's solve these equations using elimination method.
On multiplying the first equation by 7, we get

$7x+14y=35$   ...(3)

On subtracting (2) from (3), we have 

  $7x+14y=35$

$-7x-3y=-13$
_______________

  $11y=22$
 
$\Rightarrow y=2$ 

On subsituting value of y in  $x+2y=5$, we get
$x+2\left(2\right)=5$
$\Rightarrow x=1$ 

$\Rightarrow x=1$ and $y=2$
Hence, the solution of the given equations is (1, 2).