2x - y = 1
\(\Rightarrow\) y = 2x - 1

and x + 2y = 13
\(\Rightarrow \quad y=\frac { 13-x }{ 2 } \)

Plotting the above points and drawing the lines joining them, we get the graph of above equations. Clearly, two lines intersect at point A(3, 5). Hence, x = 3 and .y = 5 is the solution of above equations.
ABC is the triangular shaded region formed by the lines and the Y-axis.

We know that, in a cyclic quadrilateral, the sum of two opposite angles is 180°.
\(\therefore \quad \angle B+\angle D={ 180 }^{ 0 } and \quad \angle A+\angle C={ 180 }^{ 0 }\)
\(\Rightarrow \) 3y-5-7x+5=180 and 4y+20-4x=180
\(\Rightarrow \) 3y-7x=180   ...(i)
and 4y-4x=160 
\(\Rightarrow \) y-x=4. [dividing both sides by 4]  ...(ii)
On multiplying Eq. (ii) by 7 and then subtracting from Eq. (i), we get
-4y=180-280
\(\Rightarrow \) -4y=-100 y=25
On putting y=25 in Eq. (ii), we get
25-x=40 x=-15
On putting the values of x and y, we calculate the angles as
\(\angle A=4y+20=100+20={ 120 }^{ 0 }.\) 
\(\\ \angle B=3y-5=75-5={ 70 }^{ 0 }.\) 
\(\\ \angle C=-4x=-4(-15)={ 60 }^{ 0 }.\) 
and \( \angle D=-7x+5=105+5={ 110 }^{ 0 }.\)
Hence, the angles are \(\angle A={ 120 }^{ 0 },\angle B={ 70 }^{ 0 },\angle C={ 60 }^{ 0 }\) and \( \angle D={ 110 }^{ 0 }.\)

10 km/h, 40 km/h

Plot the lines as follows (0, 0), (4,4), (6,2)

Step 1 : Multiply Equation (1) by 2 and Equation (2) by 1 to make the coefficients of x equal. Then we get the equations as
4x + 6y = 16    (3)
4x + 6y = 7       (4)
Step 2 : Subtracting Equation (4) from Equation (3),
(4x - 4x) + (6y - 6y) = 16 - 7
i.e., 0 = 9, which is a false statemnt.
Therefore, the pair of equations has no solution.

x=3, y=4, vertices of triangle are (3,4), (-1,0) and (5,0).

Yes, it is consistent. We have, for the equation
2x + 3y - 9 = 0
a₁ = 2,b₁ = 3 and c₁= -9
and for the equation, 4x + 6y - 18 = 0
a₂ = 4, b₂ = 6 and C₂ = - 18
Here \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { 2 }{ 4 } =\frac { 1 }{ 2 } \)
\(\frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \)
and \(\frac { { c }_{ 1 } }{ { c }_{ 2 } } =\frac { -9 }{ -18 } =\frac { 1 }{ 2 } \)
It is clear that \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { { c }_{ 1 } }{ { c }_{ 2 } } \)
Hence, system is consistent and dependent.

\(ad\neq bc\Rightarrow \frac { a }{ c } \neq \frac { b }{ d } \)
Hence, the pair of given linear equations has unique solution.

Since, am = bl
\(\Rightarrow \ \ \frac { a }{ l } =\frac { b }{ m } \neq \frac { c }{ n } \)
So, ax+by=c and lx + my = n has no solution.

The pair of equations y = 0 and y = -5 has no solution.

By property of rectangle, we know that its opposite sides are of equal lengths.
i.e. DC=AB \(\Rightarrow\) x+3y=13   ...(i)
and AD=BC \(\Rightarrow\) 3x+y=7   .....(ii)
On multiplying Eq. (ii) by 3 and then subtracting Eq. (i), we get

\(\Rightarrow\) x=1
On putting x=1 in Eq. (i), we get
3y=12 \(\Rightarrow\) y=4
Hence, x=1 and y=4.

x=2, y=3

Given equation are 7x-5y=3 and 14x-10y=5
Here, a₁=7, b₁=-5, c₂=-3
and a₂=14, b₂=-10, c₂=-5
Now, \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { 7 }{ 14 } =\frac { 1 }{ 2 } ,\quad \frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { -5 }{ -10 } =\frac { 1 }{ 2 } \) and \(\frac { { c }_{ 1 } }{ { c }_{ 2 } } =\frac { 3 }{ 5 } \)
Thus, for the given equations, we have
\(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } \neq \frac { { c }_{ 1 } }{ { c }_{ 2 } } \)
So, we conclude that the two paths are parallel to each other. Hence, the paths do not cross each other. 

Let the time taken by larger pipe be x h and that taken by smaller pipe be y h, then
\(\frac { 4 }{ x } +\frac { 9 }{ y } =\frac { 1 }{ 2 } \quad \quad ...(i)\)
and \(\frac { 12 }{ x } +\frac { 12 }{ y } =1\quad \quad ...(ii)\)
Solve Eq. (i) and Eq. (ii) to get values of x and y.
Larger pipe alone fill the pool in 20 h and smaller pipe alone fill the pool in 30 h.

For parallel lines, \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } \neq \frac { { c }_{ 1 } }{ { c }_{ 2 } } \)
k=-6

For (2p- 1)x + (p - 1)y - (2p + 1) = 0
a₁ = 2p -1, b₁ = P -1 and c₁ = - (2p + 1)
and for 3x + y - 1 = 0
a₂ = 3, b₂ = 1 and c₂ = - 1
The condition for no solution is
\(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { { c }_{ 1 } }{ { c }_{ 2 } } \)
\(\Rightarrow \ \frac { 2p-1 }{ 3 } =\frac { p-1 }{ 1 } \neq \frac { 2p+1 }{ 1 } \)
By \(\frac { 2p-1 }{ 3 } =\frac { p-1 }{ 1 } \)
\(\Rightarrow\) 3p-3 = 2p - 1
\(\Rightarrow\) 3p - 2p = 3- 1
\(\therefore\)   p = 2
from  \(\frac { p-1 }{ 1 } \neq 2p+1\)
We have p - 1 \(\neq \) 2p + 1
\(\Rightarrow\)   -1-1 \(\neq \) 2p - p
\(\therefore\)   p \(\neq \) -2
from  \(\\ \frac { 2p-1 }{ 3 } \neq \frac { 2p+1 }{ 1 } \)
\(\Rightarrow\) 2p - 1\(\neq \) 6p + 3
\(\Rightarrow\)  4p  \(\neq \) -4
\(\therefore\) p \(\neq \) -1

(0,0), (0,3), (-2m³), (-2,0) and area=6 sq units

Given, system of equation is
(a+b)x-2by=5a+2b+1
or (a+b)x-2by-5a-2b-1=0  ...(i)
and 3x-y=14
or 3x-y-14=0 ......(ii)
a=5, b=1