Given, equations are
x+y=a+b   ...(i)
and ax -by=a² -b²    ...(ii)
On writing the coefficients in pattern of cross-multiplication, we get

\(\Rightarrow \frac { x }{ { a }^{ 2 }-{ b }^{ 2 }b(a+b) } =\frac { y }{ a(a+b)-{ ({ a }^{ 2 }-{ b }^{ 2 }) } } =\frac { -1 }{ -b-a } \\ \Rightarrow \quad \frac { x }{ { a }^{ 2 }+ab } =\frac { y }{ { b }^{ 2 }+ab } =\frac { 1 }{ b+a } \\ \Rightarrow \quad \underset { I }{ \frac { x }{ a(a+b) } } =\underset { II }{ \frac { y }{ b(b+a) } } =\underset { III }{ \frac { 1 }{ a+b } } \)
Taking I and III terms, we get
\(\frac { x }{ a(a+b) } =\frac { 1 }{ a+b } \quad \Rightarrow \quad x=a\quad \)
Taking II and III terms, we get
\(\frac { y }{ b(b+a) } =\frac { 1 }{ a+b } \quad \Rightarrow \quad y=b\quad \)
Hence, the required solution is x=a and y=b.

Let the numerator of the fraction be x and denominator be y. Then, the fraction is \(\frac {x}{y}.\)Now, according to condition I, we have
\(\frac { x+2 }{ y+2 } =\frac { 4 }{ 5 } \)
\(\Rightarrow\) 5x + 10 = 4y + 8 [cross-multiply both sides]
 \(\Rightarrow\)5x - 4y + 2=0    ......(i)
and according to condition II, we have
\(\frac { x-2 }{ y-2 } =\frac { 1 }{ 2 } \)       [cross-multiply both sides]
\(\Rightarrow\) 2x-8=y-4 \(\Rightarrow\) 2x-y-4=0   ...(ii)
Thus, the algebraic representation of given problem is
5x-4y=2=0 and 2x-y-4=0
To obtain graphical representation, we find atleast solutions for each equation
From Eq. (i), \(y=\frac { 5x+2 }{ 4 } ,\) Table for 5x-4y+2=0 is

So, points are A (2, 3), B( - 2, - 2) and C (6, 8) we plot these points on graph paper and join them to get a straight line representing5x - 4y + 2 = O. From Eq. (ii), y = 2x -4.
Table for 2x - y - 4 = 0 is

So, points are P(O, - 4), Q(2,0) and C(6, 8) we plot these points on graph paper and join them to get a straight line representing 2x - y - 4 =0

We find that the lines representing Eq. (i) and Eq. (ii) are inter~cting at point (6, 8).

Let the unit's place digit be x and ten's place digit be y.
Original number=10y+x
Number obtained by reversing the order of digits
=10x+y
Sum of both number=99  [given]
(10y+x)+(10x+y)=99
11x+11y=99
x+y=9  .....(i) [dividing both sides by 11]
Also, given that the digits of the numbers differ by 3.
If y>x, then y=-x=3  ..(ii)
and if y On adding Eqs. (i) and (ii), we get
2y=12
\(\Rightarrow\)  \(y= \frac {12}{2}\)=6
Then, from Eq. (i),
x=9-y=9-6=3  [\(\therefore\)y=6]
\(\therefore\)Number=10y+x=10(6)+3=63
Now, adding Eqs. (i) and (iii), we get
2x=12
x=6
Then, from Eq. (i), we get
y=9-x=9-6=3  [\(\therefore\)x=6]
\(\therefore\)Number=10y+x=10(3)+6=36
Hence, the required number 63 and 36

The given pair of linear equations is
5x-4y+8=0   ....(i)
and 7x+6y-9=0    ....(ii)
On comparing with standard form of pair of linear equations, we get
a₁=5, b₁=-4, c₁=8
and a₂=7, b₂=6, c₂=-9
Here, \(\frac { 5 }{ 7 } \neq \frac { -4 }{ 6 } \)    i.e., \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } \neq \frac { { b }_{ 1 } }{ { b }_{ 2 } } \)
So, lines (i) and (ii) are intersecting lines.

Given linear equation is 2x=7=13-x.
On substituting x=2 in the given equation, we get
LHS=2 \(\times\) 2+7=11 and RHS=13-2=11
\(\therefore\)  LHS=RHS​​​​​
Hence, x=2 is a solution of the equation.

Given, a pair of linear equations is :
x - 3y - 7 = 0             .........(i)
and 3x - 3y - 15 = 0   .........(ii) ,
On comparing eqn. (i) and (ii) with ax + by + c = 0,
a₁ = 1, b₁ = - 3, c₁ = -7
and a₂ = 3, b₂ = - 3, c₂ = -15
\(\therefore \quad \frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { 1 }{ 3 } \)
and \( \frac { { b }_{ 1 } }{ b_{ 2 } } =\frac { -3 }{ -3 } =1\)
\(\\ \therefore \quad \frac { { a }_{ 1 } }{ { a }_{ 2 } } \neq \frac { { b }_{ 1 } }{ b_{ 2 } } \)
Hence, we have a unique solution.
By cross-multiplication,
\(\frac { x }{ \left| \begin{matrix} -3 & -7 \\ -3 & -15 \end{matrix} \right| } =\frac { y }{ \left| \begin{matrix} -7 & 1 \\ -15 & 3 \end{matrix} \right| } =\frac { 1 }{ \left| \begin{matrix} 1 & -3 \\ 3 & -3 \end{matrix} \right| } \)
\(\therefore \quad \frac { x }{ \left| \begin{matrix} { b }_{ 1 } & { c }_{ 1 } \\ { b }_{ 2 } & { c }_{ 2 } \end{matrix} \right| } =\frac { y }{ \left| \begin{matrix} { c }_{ 1 } & { a }_{ 1 } \\ { c }_{ 2 } & { a }_{ 2 } \end{matrix} \right| } =\frac { z }{ \left| \begin{matrix} { a }_{ 1 } & { b }_{ 1 } \\ { a }_{ 2 } & { b }_{ 2 } \end{matrix} \right| } \)
\(\Rightarrow \quad \frac { x }{ \{ (-3)(-15)-(-3)(-7)\} } =\frac { y }{ \{ (-7)(3)-(-15)(1)\} } =\frac { 1 }{ \{ (1)(-3)-(3)(-3)\} } \)
\(\Rightarrow \quad \frac { x }{ (45-21) } =\frac { y }{ (-21+15) } =\frac { 1 }{ (-3+9) } \)
\(\Rightarrow \quad \frac { x }{ 24 } =\frac { y }{ -6 } =\frac { 1 }{ 6 } \)
\(\\ \\ \Rightarrow \quad \frac { x }{ 24 } =\frac { 1 }{ 6 } \) and \(\frac { y }{ -6 } =\frac { 1 }{ 6 } \)
\(\Rightarrow \quad x=\frac { 24 }{ 6 } =\frac { -6 }{ 6 } \)
\(\Rightarrow\) x = 4 and y = -1

Let \(\\ \frac { x }{ y } \) be the fraction, where x and yare positive integers.
Given, \(\frac { x+2 }{ y+2 } =\frac { 9 }{ 11 } and\frac { x+3 }{ y+3 } =\frac { 5 }{ 6 } \)
\(\Rightarrow\) 11 \(\times\) (x + 2) = 9 \(\times\) (y + 2)
\(\Rightarrow\) 11x + 22 = 9y + 18
\(\Rightarrow\) 11x- 9y + 4 = 0
\(\Rightarrow\) 6 \(\times\)(x + 3) = 5 \(\times\) (y + 3)
\(\Rightarrow\) 6x + 18 = 5y + 15
\(\Rightarrow\) 6x - 5y + 3 = 0
The required equations are:
11x - 9y + 4 = 0          .....(i)
and 6x - 5y + 3 = 0     .....(ii)    
From eqn. (ii), 5y = 6x + 3
\(\Rightarrow \quad y=\frac { 6x+3 }{ 5 } \)      ......(iii)
On substituting y from eqn. (iii) in eqn. (i),
\(11x-9\times \left( \frac { 6x+3 }{ 5 } \right) +4=0\quad \)
\(\Rightarrow\) 55x - 9 \(\times\)(6x + 3)+20= 0
\(\Rightarrow\) 55x - 54x - 27 + 20 = 0
\(\therefore\) x = 7
On substituting x = 7 in eqn. (iii),
\(y=\frac { 6\times 7+3 }{ 5 } \)
\(\therefore\)y = 9
Hence, the fraction \(\frac { x }{ y }\ is\ \frac { 7 }{ 9 } \)

Let the number of rows be x and number of students in each row be y, then total number of students=xy
Now, xy=(y+1)(x-2) \(\Rightarrow\) x-2y=2  ..(i)
and xy=(y-1)(x+3) \(\Rightarrow\) -x+3y=3  ...(ii)
On solving Eq. (i) and Eq. (ii), we get
x=12 and y=5
Total number of students=12 x 5=60

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

Pair of linear equations kx - 4y - 3 = 0 and 6x - 12y - 9 = 0
Condition for infinite solutions: - \(\frac { { a }_{ 1 } }{ a_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { { c }_{ 1 } }{ { c }_{ 2 } } \)
\(\frac { -k }{ 6 } =\frac { -4 }{ -12 } =\frac { 3 }{ 9 } \)
\(\Rightarrow\) k = 2

The given equations can be rewritten as
3x + 2y - 5 = 0 and 2x - 3y - 7 = 0
On comparing with standard form of pair of linear equations, we get a₁ = 3, b₁ = 2, c₁ = -5
and a₂ = 2, b₂ = -3, c₂ = -7
Now, \(\frac{a_{1}}{a_{2}}=\frac{3}{2}, \frac{b_{1}}{b_{2}}=-\frac{2}{3}\) and \(\frac{c_{1}}{c_{2}}=\frac{5}{7}\)
Thus, \(\frac{3}{2}\neq -\frac{2}{3},i.e.\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}\)
Hence, the pair of linear equations is consistent.

The equation of one line is 4x + 3y = 14. We know that if two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are parallel, then
\(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } \neq \frac { { c }_{ 1 } }{ c_{ 2 } } \)
\(\Rightarrow \ \ \ \frac { 4 }{ { a }_{ 2 } } =\frac { 3 }{ { b }_{ 2 } } \neq \frac { { c }_{ 1 } }{ { c }_{ 2 } } \Rightarrow \frac { { a }_{ 2 } }{ { b }_{ 2 } } =\frac { 4 }{ 3 } =\frac { 12 }{ 9 } \)
Hence one of the possible, second parallel line is 12x + 9y = 5

\(ad\neq bc\Rightarrow \frac { a }{ c } \neq \frac { b }{ d } \)
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.

x=-1, y=3

x=4, y=-1

k=-3/2

Let father's age be x yr and sum of ages of his two children be y yr.
For condition I
x=3y   ..(i)
For condition II
After 5 yr, age of father=(x+5)yr
After 5 yr, sum of ages of two children=(y+10) yr
x+5=2(y+10)   ..(ii)
On putting the value of x from Eq. (i) in Eq. (ii), we get
3y+5=2(y+10)
\(\Rightarrow\)   3y+5=2y+20
\(\Rightarrow\) 3y+2y=20-5
\(\Rightarrow\)   y=15
On putting the value of y in Eq. (i), we get
x=3(15)=45
Hence, the age of father is 45 yr.

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