The given system of equations are:
(a - b)x + (a + b)y = 2a² - 2b²
So, (a - b)x + (a + b)y - 2a² - 2b² = 0
 (a - b)x + (a + b)y - 2(a² - b²) = 0 ........(i)
And (a + b)(x + y) = 4ab
So, (a +b)x + (a + b)y - 4ab = 0 ..........(ii)
The given system of equation is in the form of
a₁x + b₁y - c₁ = 0
and a₂x + b₂y - c₂ = 0
Compare (i) and (ii) , we get
a₁ = a - b, b₁ = a + b, c₁ = -2(a² + b²)
a₂ = a + b, b₂ = a + b, c₂ = -4ab
By cross-multiplication method
{tex}\frac{x}{{2(a + b)({a^2} - {b^2} + 2ab)}}{/tex} {tex} = \frac{{ - y}}{{2(a - b)({a^2} + {b^2})}}{/tex} {tex} = \frac{1}{{ - 2b(a + b)}}{/tex}
Now, {tex}\frac{x}{{2(a + b)({a^2} - {b^2} + 2ab)}} = \frac{1}{{ - 2b(a + b)}}{/tex} {tex}{/tex}

{tex}⇒ x = \frac{{2ab - {a^2} + {b^2}}}{b}{/tex}
And, {tex}\frac{{ - y}}{{2(a - b)({a^2} + {b^2})}} = \frac{1}{{ - 2b(a + b)}} {/tex} 
{tex}⇒ y = \frac{{(a - b)({a^2} - {b^2})}}{{b(a + b)}}{/tex}
The solution of the system of equations are {tex}\frac{{2ab - {a^2} + {b^2}}}{b}{/tex} and {tex}\frac{{(a - b)({a^2} - {b^2})}}{{b(a + b)}}{/tex} respectively.

Given linear equation is
(3k + 1)x + 3 y - 2 = 0 .......... (i)
(k² + 1)x + (k - 2)y - 5 = 0 ............ (ii)
Compare with a₁x + b₁y + c = 0 and a₂x + b₂y and c₂ = 0
a₁ = 3k + 1 , b₁ = 3 , c₁ = -2
and a₂ = k²+ 1 , b₂ = k - 2, c₂ = -5
The given system of equations will have no solution, if
{tex} \frac { a_1 } { a_2 } = \frac { b_1 } { b_2} \neq \frac { c_1 } { c_2 }{/tex}
{tex} \frac { 3 k + 1 } { k ^ { 2 } + 1 } = \frac { 3 } { k - 2 } \neq \frac { - 2 } { - 5 }{/tex}
{tex}\Rightarrow \quad \frac { 3 k + 1 } { k ^ { 2 } + 1 } = \frac { 3 } { k - 2 } \text { and } \frac { 3 } { k - 2 } \neq \frac { 2 } { 5 }{/tex}
Now, {tex}\frac { 3 k + 1 } { k ^ { 2 } + 1 } = \frac { 3 } { k - 2 }{/tex}
{tex}\Rightarrow{/tex} (3k + 1)(k - 2)=3(k² + 1)
{tex}\Rightarrow{/tex} 3k² - 5k - 2 =3k² + 3
{tex}\Rightarrow{/tex} -5k - 2 =3
{tex}\Rightarrow{/tex} -5k = 5
{tex}\Rightarrow{/tex} k = -1
Clearly, {tex}\frac { 3 } { k - 2 } \neq \frac { 2 } { 5 }{/tex} for k = -1.
Hence, the given system of equations will have no solution for k = -1.

We do not know whether {tex} \frac { a } { b } < \frac { a + 2 b } { a + b } \text { or, } \frac { a } { b } > \frac { a + 2 b } { a + b }{/tex}.
Therefore, to compare these two numbers, let us compute {tex} \frac { a } { b } - \frac { a + 2 b } { a + b }{/tex}
We have,
{tex} \frac { a } { b } - \frac { a + 2 b } { a + b } = \frac { a ( a + b ) - b ( a + 2 b ) } { b ( a + b ) }{/tex} {tex} = \frac { a ^ { 2 } + a b - a b - 2 b ^ { 2 } } { b ( a + b ) } = \frac { a ^ { 2 } - 2 b ^ { 2 } } { b ( a + b ) }{/tex}
{tex} \therefore \quad \frac { a } { b } - \frac { a + 2 b } { a + b } > 0{/tex}
{tex} \Rightarrow \quad \frac { a ^ { 2 } - 2 b ^ { 2 } } { b ( a + b ) } > 0{/tex}
{tex} \Rightarrow{/tex}    a² - 2b² > 0
{tex} \Rightarrow{/tex} a²> 2b²
{tex} \Rightarrow \quad a > \sqrt { 2 } b{/tex}
and, {tex} \frac { a } { b } - \frac { a + 2 b } { a + b } < 0{/tex}
{tex} \Rightarrow \quad \frac { a ^ { 2 } - 2 b ^ { 2 } } { b ( a + b ) } < 0{/tex}   
{tex} \Rightarrow{/tex} a² - 2b² < 0
{tex} \Rightarrow{/tex}a² <2b²
{tex} \Rightarrow \quad a < \sqrt { 2 } b{/tex}
Thus, {tex} \frac { a } { b } > \frac { a + 2 b } { a + b }{/tex}, if {tex}a > \sqrt { 2 b }{/tex} and {tex} \frac { a } { b } < \frac { a + 2 b } { a + b }{/tex}, if {tex} a < \sqrt { 2 } b{/tex}.
So, we have the following cases:
CASE I When {tex} a > \sqrt { 2 } b{/tex}
In this case, we have
{tex} \frac { a } { b } > \frac { a + 2 b } { a + b } \text { i.e., } \frac { a + 2 b } { a + b } < \frac { a } { b }{/tex}
We have to prove that
{tex} \frac { a + 2 b } { a + b } < \sqrt { 2 } < \frac { a } { b }{/tex}
We have,
{tex} a > \sqrt { 2 } b{/tex}
{tex} \Rightarrow{/tex} a²> 2b²  [Adding a² on both sides]
{tex} \Rightarrow \quad 2 a ^ { 2 } + 2 b ^ { 2 } > \left( a ^ { 2 } + 2 b ^ { 2 } \right) + 2 b ^ { 2 }{/tex} [Adding 2b² on both sides]
{tex} \Rightarrow \quad 2 \left( a ^ { 2 } + b ^ { 2 } \right) + 4 a b > a ^ { 2 } + 4 b ^ { 2 } + 4 a b{/tex} [Adding 4ab on both sides]
{tex} \Rightarrow \quad 2 \left( a ^ { 2 } + 2 a b + b ^ { 2 } \right) > a ^ { 2 } + 4 a b + 4 b ^ { 2 }{/tex}
{tex} \Rightarrow \quad 2 ( a + b ) ^ { 2 } > ( a + 2 b ) ^ { 2 }{/tex}
{tex} \Rightarrow \quad \sqrt { 2 } ( a + b ) > a + 2 b{/tex}
{tex} \Rightarrow \quad \sqrt { 2 } > \frac { a + 2 b } { a + b }{/tex}  ........(i)
Again,
{tex} a > \sqrt { 2 } b {/tex}
{tex}\Rightarrow \frac { a } { b } > \sqrt { 2 }{/tex}  .......(ii)
From (i) and (ii), we get
{tex} \frac { a + 2 b } { a + b } < \sqrt { 2 } < \frac { a } { b }{/tex}
CASE II When {tex} a < \sqrt { 2 } b{/tex}
In this case, we have
{tex} \frac { a } { b } < \frac { a + 2 b } { a + b }{/tex}
We have to show that {tex} \frac { a } { b } < \sqrt { 2 } < \frac { a + 2 b } { a + b }{/tex}
We have,
{tex} a < \sqrt { 2 } b{/tex}
{tex} \Rightarrow \quad a ^ { 2 } < 2 b ^ { 2 }{/tex}
{tex} \Rightarrow \quad a ^ { 2 } + a ^ { 2 } < a ^ { 2 } + 2 b ^ { 2 }{/tex} [Adding a² on both sides]
{tex} \Rightarrow \quad 2 a ^ { 2 } + 2 b ^ { 2 } < a ^ { 2 } + 2 b ^ { 2 }+ 2 b ^ { 2 }{/tex} [Adding 2b² on both sides]
{tex}\Rightarrow \quad 2 a ^ { 2 } + 2 b ^ { 2 } < a ^ { 2 } + 4 b ^ { 2 }{/tex}
{tex} \Rightarrow \quad 2 a ^ { 2 } + 4 a b + 2 b ^ { 2 } < a ^ { 2 } + 4 a b + 4 b ^ { 2 }{/tex} [Adding 4ab on both sides]
{tex} \Rightarrow \quad 2 ( a + b ) ^ { 2 } < ( a + 2 b ) ^ { 2 }{/tex}
{tex} \Rightarrow \sqrt { 2 } ( a + b ) < a + 2 b{/tex}
{tex} \Rightarrow \quad \sqrt { 2 } < \frac { a + 2 b } { a + b }{/tex} . ...(iii)
{tex} \Rightarrow \quad a < \sqrt { 2 } b \Rightarrow \frac { a } { b } < \sqrt { 2 }{/tex}  ....(iv)
From (iii) and (iv), we get
{tex} \frac { a } { b } < \sqrt { 2 } < \frac { a + 2 b } { a + b }{/tex}
Hence, {tex} \sqrt { 2 }{/tex} lies between {tex} \frac { a } { b }{/tex} and {tex} \frac { a + 2 b } { a + b }{/tex}.

Taking different values of n, we find that A and B are co-prime.
When {tex}n = 1, A = 5, B = 8{/tex}
When {tex}n = 2, A = 7, B = 9{/tex}
When {tex}n = 3, A = 9, B = 10{/tex} and so on....
Therefore, {tex}HCF = 1{/tex}

HCF of 56 and 72

{tex}\begin{array}{l}56=8\times7=2^3\times7\\72=8\times9=2^3\times3^2\\So\;HCF(56,72)=2^3=8\end{array}{/tex}

d = 56x + 72y

⇒ 8 = 56x + 72y

Dividing by 8 both sides

1= 7x + 9y

Put x = 4 and y = –3

1 = 7 × 4 + 9(–3)

= 28 – 27

1 = 1

L.H.S = R.H.S.

Put x = –5 and y = 4

1 = 7(–5) + 9 {tex} \times {/tex} 4

= –35 + 36

1 = 1

L.H.S = R.H.S.

x = 4, and y = –3
x = –5 and y = 4
Satisfy the equations
 ∴  x and y are not unique.

{tex}\frac { 10 } { x + y } + \frac { 2 } { x - y } = 4{/tex}
{tex}\frac { 15 } { x + y } - \frac { 9 } { x - y } = - 2{/tex}.
Putting {tex}\frac 1{x+y}{/tex}=u and {tex}\frac 1{x-y}{/tex}= v ,the given equations
{tex}10u + 2v= 4{/tex}........(i)
{tex}15u - 9v = -2{/tex}.......(ii)

Multiplying (i) by 9 and (ii) by 2 and adding them,
{tex}\Rightarrow{/tex} {tex}90 u + 18 v = 36\ and\ 30u - 18v = -4{/tex}
{tex}\Rightarrow 120u = 32{/tex}
{tex}\Rightarrow u = \frac{4}{15}{/tex}
Substituting u = {tex}\frac{4}{15}{/tex} in (i), we get v = {tex}\frac { 2 } { 3 }{/tex}
{tex}\Rightarrow \frac { 1 } { x + y } = \frac { 4 } { 15 } \text { and } \frac { 1 } { x - y } = \frac { 2 } { 3 }{/tex}
{tex}\Rightarrow x + y = \frac { 15 } { 4 }{/tex} .......(iii)
and {tex}x - y = \frac { 3 } { 2 }{/tex} ......(iv)
Adding (iii) and (iv), 
{tex}\Rightarrow 2 x = \frac { 21 } { 4 } \Rightarrow x = \frac { 21 } { 8 }{/tex}
Substituting x ={tex}\frac { 21 } { 8}{/tex} in (iii), we get {tex}y = \frac { 9 } { 8 }{/tex}
Hence, the solution is {tex}x = \frac { 21 } { 8 } \text { and } v = \frac { 9 } { 8 }{/tex}

Given numbers are 657 and 963 .
Here, 657 < 963 
By using Euclid's Division algorithmm , we get
963 = (657  × 1) + 306
Here , remainder = 306 .
So, On taking 657 as new dividend and 306 as the new divisor and then apply Euclid's Division lemma, we get
657 = (306  × 2) + 45

Here, remainder = 45 

So, On taking 306 as new dividend and 45 as the new divisor and then apply Euclid's Division lemma, we get
306 = (45  × 6) + 36

Here, remainder = 36

So, On taking 45 as new dividend and 36 as the new divisor and then apply Euclid's Division lemma, we get
45 = (36  × 1) + 9
Here, remainder = 9

So, On taking 36 as new dividend and 9 as the new divisor and then apply Euclid's Division lemma, we get

36 = (9  × 4) + 0
Here , remainder = 0 and last divisor is 9. 

Hence, HCF of  657 and 963 = 9.

∴ 9 = 657x + 963(-15)
⇒ 9 = 657x - 14445
⇒ 657x = 9 + 14445
⇒ 657x = 14454
⇒x = 14454/657
⇒ x =22