The given system of equations are
3x − 4y = 10 \(\Rightarrow\) 3x − 4y −10 = 0 .....(1)
  4x + 3y = 5 \(\Rightarrow\) 4x + 3y − 5 = 0 .....(2)
For the cross multiplication method, we write the coefficients as

\(\frac { x }{ (-4)(-5)-(3)(-10) } =\frac { y }{ (-10)(4)-(-5)(3) } =\frac { 1 }{ (3)(3)-(4)(-4) } \)
\(\frac { x }{ (20)-(-30) } =\frac { y }{ (-40)-(-15) } =\frac { 1 }{ (9)-(-16) } \)
\(\frac { x }{ 20+30 } =\frac { y }{ -40+15 } =\frac { 1 }{ 9+16 } \)
\(\frac { x }{ 50 } =\frac { y }{ -25 } =\frac { 1 }{ 25 } \)
Therefore, we get \(x=\frac { 50 }{ 25 } ;y=\frac { -25 }{ 25 } \)
x = 2; y = -1
Thus the solution is x = 2, y = –1.
Verification:
3x–4y = 10 ...(1)
3(2)–4(–1) = 10
6 + 4 = 10
10 = 10 True
4x + 3y = 5 ...(2)
4(2) + 3(–1) = 5
8–3 = 5
5 = 5 True

(i) 2x - y = 3    ....(1)
3x + y = 7        ...(2)
(1)+(2) \(\Rightarrow\) 2x - y = 3
\(\cfrac { 3x+y=7 }{ 5x=10 } \)
\(x=\cfrac { 10 }{ 5 } =2\)
Substitute x = 2 in (1)
2(2) -y = 3
4-y = 3 
-y = 3 - 4
-y = -1
\(\therefore\) x = 2; y = 1
(ii) x - y = 5                  ....(1)
3x + 2y = 25                ...(2)
(1) \(\times\) 3 \(\Rightarrow\) 3x - 3y = 15      ...(3)
(2) \(\Rightarrow \cfrac { 3x+2y=25 }{ -5y=-10 } \)
(3) - (2)
y = 2
Substitute y = 2 in (1)
x - 2 = 5
x = 5 + 2
x = 7
\(\therefore\) x = 7, y = 2
(iii) \(\cfrac { x }{ 10 } +\cfrac { y }{ 5 } =14\Rightarrow \cfrac { x+2y }{ 10 } =14\)
x + 2y = 140    .......(1)
\(\cfrac { x }{ 10 } +\cfrac { y }{ 6 } =15\Rightarrow \cfrac { 3x+4y }{ 24 } =15\)
3x + 4y = 360    ......... (2)

Substitute y = 30 in (1)
x + 2 (30) = 140
x + 60 = 140
x = 140 - 60
x = 80
\(\therefore\) x = 80; y = 30
(iv) 3(2x + y) = 7xy \(\Rightarrow\) 6x + 3y = 7xy
3(x + 3y) = 11xy \(\Rightarrow\) 3x + 9y = 11xy

\(\cfrac { 6 }{ y } +\cfrac { 3 }{ x } =7\)

\(\cfrac { 3 }{ y } +\cfrac { 9 }{ x } =11\)      ....(4)
In (3), (4) Put \(\cfrac { 1 }{ x } =a,\cfrac { 1 }{ y } =b\)
(3) \(\Rightarrow\) 36b + 3a = 7
(4) \(\Rightarrow\) 3b +9a = 11
(6) X (2) \(\Rightarrow\) 6b + 18a = 22 
(5) \(\Rightarrow\) \(\cfrac { 6b+3a=7 }{ 15a=15 } \)
(3)-(2)
\(a=\cfrac { 15 }{ 15 } \)
Substitute a = 1 in (5)
6b + 3 (1) = 7
6b + 3 = 7
6b = 7 - 3
\(b=\cfrac { 4 }{ 6 } =\cfrac { 2 }{ 3 } \)
\(\therefore \ a=\cfrac { 1 }{ x } =1\Rightarrow x=1\)
\(b=\cfrac { 1 }{ y } =\cfrac { 2 }{ 3 } \Rightarrow y=\cfrac { 3 }{ 2 } \)
\(\therefore\) Solution x = 1; \(y=\cfrac { 3 }{ 2 } \)
(iv) Put  \(\cfrac { 1 }{ x } =a\)
\(\cfrac { 4 }{ x } +5y=7\Rightarrow 4a+5y=7\)   ....(1)
\(\cfrac { 3 }{ x } +4y=5\Rightarrow 3a+4y=5\)    ...(2)
(1) \(\times\) 3 \(\Rightarrow\) 12a +15y = 21
(2)  4 \(\Rightarrow\) \(\cfrac { 12a+16y=20 }{ -1y=1 } \)
y = 1 
Substitute y = -1 in (1)
4a + 5 (-1)=7
4a - 5 = 7
4a = 7 + 5
\(a=\cfrac { 12 }{ 4 } =3\)
\(a=\cfrac { 1 }{ x } =3\Rightarrow x=\cfrac { 1 }{ 3 } \)
\(\therefore\) y = -1
\(\therefore\)  \(x=\cfrac { 1 }{ 3 } \) ; y = -1
(vi) 13x + 11y = 70 --------- (1)
11x + 13y = 74 --------- (2)
(1) \(\times\) 11 \(\Rightarrow\) 143x+121y = 770   .... (3)
(2) \(\times\) 13 \(\Rightarrow\) \(\cfrac { 143x+169y=962 }{ -48y=-192 } \)  ....(4)
(3) - (4) 
\(y=\cfrac { 192 }{ 48 } =4\)
Substitute y = 4 in (1) 48
13x + 11 (4) = 70
13x + 44 = 70
13x = 70 - 44 = 26
\(x=\cfrac { 26 }{ 13 } =2\)
\(\therefore\)  x = 2; y = 4 

(i)  2x- 3y = 7   ...(1)
5x + y = 9
Step (1): From the equation (2)
5x + y = 9
y = -5x + 9  ..(3)
Step (2): substitute (3) in (1)
2x - 3(-5x + 9) = 7
2x + 15x -27 = 7
17x = 7 + 27
17x = 34
\(x=\cfrac { 34 }{ 17 } =2;x=2\)
Step (3): substitute x = 2 in (3)
y = -5(2) + 9
= 10 + 9 = -1
x = 2
y = -1
(ii) 1.5x + 0.1y = 6.2    ...(1)
3x - 0.4y = 11.2  ...(2)
Multiply (1)\(\times\) 10 \(\Rightarrow\)15x + y = 62  ...(1)
(2) \(\times\) 10 \(\Rightarrow\)30x - 4y = 112   ....(4)
Step (1): From equation (3)
15x+ y = 62
y = -15x + 62  ...(5)
Step (2): substitute (5) in (4)
30x - 4 (-15x + 62) =112
30x + 60x -248 = 112
90x = 112 + 248
90x = 360

Step (3): substitute x = 4  in (5)
y = -15(4) + 62
= -60 + 62 = 2
x = 4
y = 2
(iii) 10% of x + 20% of y = 24; 3x - y = 20

\(\cfrac { x }{ 10 } +\cfrac { y }{ 5 } =24\)
\(\cfrac { x+2y }{ 10 } =24\)
x + 2y = 240    ....(1)
3x-y = 20         ....(2)
Step (1):  From equation (2)
Step (2): substitute (3) in (1) 
x + 2 (3x - 20) = 240
x + 6x - 40 = 240
7x = 240 + 40
\(x=\cfrac { 280 }{ 7 } \)
x = 40
Step (3): substitute x = 40 in (3)
y = 3 (40) - 20
= 120 - 20 = 100
x = 40
y = 100
(iv) \(\sqrt { 2x } -\sqrt { 3y } =1\)    ...(1)
\(\sqrt { 3x } -\sqrt { 8y } =0\)          ..(2)
From the equation (2)
\(\sqrt { 3x } -\sqrt { 8y } =0\)

\(y=\ \cfrac { \sqrt { 3 } }{ \sqrt { 8 } } x\)   ...(3)
Step (2): substitute (3) in (1)
\(\sqrt { 2x } -\sqrt { 3y } =1\)
\(\sqrt { 2x } -\sqrt { 3 } \left( \cfrac { \sqrt { 3 } }{ \sqrt { 8 } } \right) x=1\)
\(\cfrac { \sqrt { 2x } }{ 1 } -\cfrac { 3 }{ \sqrt { 8 } } x=1\)
\(\cfrac { \sqrt { 16x } -3x }{ \sqrt { 8 } } =\cfrac { 4x-3x }{ \sqrt { 8 } } =1\quad x=\sqrt { 8 } \)
Step (3): substitute x = \(\sqrt { 8 } \) in (1)
\(\sqrt { 2 } \left( \sqrt { 8 } \right) -\sqrt { 3y } =1\)
\(\sqrt { 16 } -\sqrt { 3y } =1\)
\(4-\sqrt { 3y } =1\)
\(-\sqrt { 3y } =1-4\)
 
\(x=\sqrt { 8 } \) 
\(y=\sqrt { 3 } \) 

Let x be the digit at ten’s place and y be the digit at unit place.
Given that x + y = 5 …… (1)

Given, Original number − reversing number = 27
(10x + y) − (10y + x) = 27
10x − x + y −10y = 27
9x − 9y = 27
\(\Rightarrow\) x − y = 3 ... (2)
Also from (1), y = 5 – x ... (3)
Substitute (3) in (2) to get x − (5 − x) = 3
x − 5 + x = 3
2x = 8
x = 4
Substituting x = 4 in (3), we get y = 5 − x = 5 − 4
y = 1
Thus, 10x + y = 10 × 4 +1 = 40 +1 = 41.
Therefore, the given two-digit number is 41.
Verification :
sum of the digits = 5
x + y = 5
4 + 1 = 5
5 = 5 true
Original number – reversed number = 27
41 - 14 = 27
27 = 27 true

Let us form table of values for each line and then fix the ordered pairs to be Plotted.
Graph of y = 7 - x

Points to be plotted : (-2, 9), (-1, 8), (0, 7), (1, 6), (2, 5)
Graph of y = x - 3

Points to be plotted (-2, -5), (-1, -4), (0, -3), (1, -2), (2, -1) 

Let us form table of values for each line and then fix the ordered pairs to be plotted.
Graph of 3x + 2y = 6

Points to be plotted:
(−2,6), (0,3), (2,0)
Graph of 6x + 4y = 8

Points to be plotted:
(-2, 5), (0, 2), (2, -1)
When we draw the graphs of these two equations, we find that they are parallel and they fail to meet to give a point of intersection. As a result there is no ordered pair that can be common to both the equations. In this case there is no solution to the system.

Given x + y = 5 ...(1)
2x – y = 4 ...(2)
To draw the graph (1) is very easy. We can find the x and y intercepts and thus two of the points on the line (1).
When x = 0, (1) gives y = 5.
Thus A(0,5) is a point on the line.
When y = 0, (1) gives x = 5.
Thus B(5,0) is another point on the line.
Plot A and B; join them to produce the line (1).
To draw the graph of (2), we can adopt the same procedure.
When x = 0, (2) gives y = −4.
Thus P(0,−4) is a point on the line.
When y = 0, (2) gives x = 2.
Thus Q(2,0) is another point on the line.
Plot P and Q; join them to produce the line (2).
The point of intersection (3, 2) of lines (1) and (2) is a solution.
The solution is the point that is common to both the lines. Here we find it to be (3,2). We can give the solution as x = 3 and y = 2.

(i) Let us prepare a table to find the ordered pairs of points for the line y = 3x −1.
We shall assume any value for x, for our convenience let us take −1, 0 and 1.
When x = −1, y = 3(–1)–1 = –4
When x = 0, y = 3(0)–1 = –1
When x = 1, y = 3(1)–1 = 2

The points (x,y) to be plotted :
(−1, −4), (0, −1) and (1, 2).

(ii) Let us prepare a table to find the ordered pairs of points for the line y =\(\left( \frac { 2 }{ 3 } \right) x+3\)
Let us assume −3, 0, 3 as x values.
(why?)
When x = -3, \(y=\frac { 2 }{ 3 } (-3)+3=1\)
When x = 0, \(y=\frac { 2 }{ 3 } (0)+3=3\)
When x =3, \(y=\frac { 2 }{ 3 } (3)+3=5\)

The points (x, y) to be plotted: (-3, 1), (0, 3) and (3, 5).

We have already come across one method: forming a table of values, listing and plotting ordered pairs and joining the points.
But, to fix a line, after all, how many points do we need? Just two! These can easily be obtained when a line is given in the form y = mx + c.
The given line y = 4x – 3
put x = 0 to get y-intercept
y = 4(0)–3
y = –3
point is (0, –3) and y-intercept = – 3
put y = 0 to get x-intercept
0 = 4x – 3
3 = 4x
\(\frac { 3 }{ 4 } =x\)
point is \(\left( \frac { 3 }{ 4 } ,0 \right) \) and x-intercept = \(\frac { 3 }{ 4 } \)
The graph may be drawn through two points (0,−3) and \(\left( \frac { 3 }{ 4 } ,0 \right) \)