Ax by is equal to a minus bbx minus y is equal to a + b solve by cross multiplication method

The system of the given equations may be written as

ax + by - a2 = 0

bx + ay - b2 = 0

Here,

`a_1 = a, b_1 = b, c_1 = -a^2`

`a_2 = b, b_2 = a, c_2 = -b^2`

By cross multiplication, we get

`=> x/(b xx (-b^2) - (-a^2) xx a) = (-y)/(a xx(-b^2) - (-a^2) xx b) = 1/(axxa - bxxb)`

`=> x/(-b^3 + a^3) = (-y)/(-ab^2 + a^2b) = 1/(a^2 - b^2)`

Now

`x/(-b^3 + a^3) = 1/(a^2 - b^2)`

`=> x = (a^3 - b^3)/(a^2 - b^2)`

`= ((a- b)(a^2 + ab + b^2))/((a- b)(a + b))`

`= (a^2 + ab + b^2)/(a + b)`

And

`(-y)/(-ab^2 + a^2b) = 1/(a^2 - b^2)`

`=> -y = (a^2b - ab^2)/(a^2 - b^2)`

`=> y = (ab^2 - a^2b)/(a^2 - b^2)`

`=> (ab(b -a))/((a-b)(a + b))`

`(-ab(a - b))/((a - b)(a + b))`

`= (-ab)/(a + b)`

Hence `x = (a^2 + ab + b^2)/(a + b), y = (-ab)/(a + b)` is the solution of the given system of the equations.

Page 2

Let `1/(x + y) = u` and `1/(x - y) = v` Then given system of equations becomes

5u - 2v = -1

15u + 7v = 10

here

`a_1 = 5, b_1 = -2, c_1 = 1`

`a_2 = 15, b_2 = 7,c_2 = -10`

By cross multiplication, we get

`=> u/((-2)xx (-10)-1 xx7) = u/(5 xx (-10)-1 xx 15) = 1/(5xx7 - (-2) xx 15 )`

`=> u/(20 - 7) = (-v)/(-50 - 15) = 1/(35 + 30)`

`=> u/13 = (-v)/(-65) = 1/65`

`=> u/13 = v/65 = 1/65`

Now

`u/13 = 1/65`

`=> u = 13/65 = 1/5`

And

`v/65 = 1/65`

`=> v = 65/65 = 1`

Now

`u = 1/(x + y)`

`=> 1/(x + y) = 1/5`    ....(i)

And

`v = 1/(x - y)`

`=> 1/(x - y) = 1`

=> x - y = 1 .....(ii)

Adding equation (i) and (ii), we get

2x = 5 + 1

=> 2x = 6

`=> x = 6/2 = 3`

Adding x = 3 in eq 2

3 - y = 1

y = 3 -1

y = 2

Page 3

The given system of equation is

``2/x + 3/y = 13`

`5/x - 4/y = -2 " where " x!= 0 and  y != 0`

Let `1/x = u` and `1/y = v`, Then the given system of equation becomes

2u + 3v = 13

5u - 4v = -2

here

`a_1 = 2, b_1 = 3, c_1 = -13`

`a_2 = 5, b_2 = -4, c_2 = 2`

By cross multiplication, we have

`=> u/(3xx2-(-13)xx(-4)) = (-v)/(2xx2-(-13)xx5) = 1/(2xx (-4)-3 xx 5)`

`=> u/(6 - 52) = (-v)/(4 + 65) = 1/(-8 - 15)`

`=> u/(-46) = (-v)/69 = 1/(-23)`

Now

`u/(-46) = 1/(-23)`

`=> u = (-46)/(-23) = 2`

And

`(-v)/69 =1/(-23)`

`=> v = (-69)/(-23) = 3`

Now

`x = 1/u = 1/2`

And

`y = 1/v = 1/3``

Hence `x = 1/2, y = 1/3` is the solution of the given system of equations.

Page 4

`x/a = y/b`

``ax + by = a^2 + b^2`

Here `a_1 = 1/a, b_1 = (-1)/b, c_1 = 0`

`a_2 = a, b_2 = b,c_2 = -(a^2 + b^2)`

By cross multiplication, we get

`x/(-1/b(-(a^2 + b^2))-b(0)) = (-y)/(1/a(-(a^2 + b^2))-a(0)) = 1/(1/a (b) - a xx ((-1)/b))`

`x/((a^2 + b^2)/b) = y/((a^2 + b^2)/a) = 1/(b/a + a/b)`

`x = ((a^2 + b^2)/b)/(b/a + a/b) = ((a^2 +b^2)/b)/((b^2 + a^2)/(ab)) = a`

`y = ((a^2 + b^2)/a)/(b/a + a/b) = ((a^2 + b^2)/b)/((b^2+a^2)/(ab)) = b`

Solution is (a, b)

Page 5

The system of the given equations may be written as

`1/a x xx + 1/b xx y - 2 = 0`

`ax - by + b^2 - a^2 = 0`

here

`a_1 = 1/a, b_1 = 1/b, c_1 = -2`

`a_2 = a, b_2= -b, c_2 = b^2 - a^2`

By cross multiplication, we get

`=> x/(1/b xx (b^2 - a^2) - (-2) xx (-b)) = (-y)/(1/a xx (b^2 - a^2) - (-2) xx a) = 1/((-bxx1)/a - (a xx1)/b)`

`=> x/((b^2 - a^2)/b - 2b) = (-y)/((b^2 - a^2)/b + 2b) = 1/((-b)/a - a/b)`

`=> x/((b^2 -a^2 - 2b^2)/b) = (-y)/((b^2 - a^2 + 2b^2)/a) = 1/((-b^2 - a^2)/(ab)`

`=> x/((a^2 - b^2)/b) = (-y)/((b^2 + a^2)/a) = 1/((-b^2 -a^2)/(ab)`

Now

`x/((-a^2 -b^2)/b) = 1/((-b^2 - a^2)/(ab)`

`=> x = (-a^2 - b^2)/b xx (ab)/(-b^2 - a^2)`

And

`(-y)/((b^2 + a^2)/a)= 1/((-b^2 -a^2)/(ab))`

`=> -y = (b^2 + a^2)/a xx (ab)/(-b^2 - a^2)`

`=> -y = ((b^2 + a^2)xxb)/(-(b^2 + a^2)`

=> y = b

Hence, , x = a, y = b is the solution of the given system of the equations.

Page 6

The given system of equation is

`ax + by = (a + b)/2` .....(i)

3x + 5y = 4  ....(ii)

From (i), we get

2(ax + by) = a + b

`=> 2ax + 2by - (a + b) = 0`  .....(iii)

From (ii), we get

3x + 5y - 4 = 09

here

`a_1 = 2a, b_1 = 2b, c_1 = -(a + b)`

`a_2 = 3, b_2 = 5, c_2 = -4`

By cross multiplication, we hav-04e

`=> x/(2b xx (-4)- [-(a + b)]xx5) = (-y)/(2a xx (-4) - [-(a + b)] xx 3) = 1/(2a xx 5 - 2b xx 3) `

`=> x/(-8b + 5(a + b)) = (-y)/(-8a + 3(a + b)) = 1/(10a - 6b)`

`=> x/(-8b + 5a + 5b) = (-y)/(-8a + 3a + 3b) = 1/(10a - 6b)` 

`=> x/(5a - 3b) = (-y)/(-5a + 3b) = 1/(10a - 6b)`

Now

`x/(5a - 3b) = (-y)/(-5a + 3b) = 1/(10a - 6b)`

`=> x = (5a -3b)/(10a - 6b) = (5a - 3b)/(2(5a - 3b)) = 1/2`

And

`(-y)/(-5a + 3b) = 1/(10a - 6b)`

`=> -y = (-5a + 3b)/(2(5a - 3b))`

`=> y = (-(-5a + 3b))/(2(5a - 3b))`

`= (5a - 3b)/(2(5a - 3b))`

`=> y = 1/2`

Hence x = 1/2, y = 1/2  is the solution of the given system of equations

Page 7

The given system of equations is

2ax + 3by = a + 2b ....(i)

3ax + 2by = 2a + b  ....(ii)

Here

`a_1 = 2a, b_1 = 3b, c_1 = -(a + 2b)`

`a_2 = 3z, b_2 = 2b, c_2 = -(2a + b) `

By cross multiplication we have

`=> x/(-3b xx (2a + b) - [-(a + 2b)]xx2b) = (-y)/(-2a xx (2a + b)-[-(a +2b)] xx 3a) = 1/(2a xx 2b - 3b xx 3a)`

`=> x/(-3b + (2a + b) + 2b (a + 2b)) = (-y)/(-2a(2a + b) + 3a (a + 2b)) = 1/(4ab - 9ab)`

`=> x/(-6ab = 3b^2  + 2ab + 4b^2) = (-y)/(-4a^2 -2ab
+ 3a^2 + 6ab) = 1/(4ab - 9ab)`

`=> x/(-4ab + b^2) = (-y)/(-a^2 + 4ab) = 1/(-5ab)`

Now

`x/(-4ab + b^2) =- 1/(-5ab)`

`=> x = (-4ab + b^2)/(-5ab)`

`=> (-b(4a - b))/(-5ab)`

`= (4a - b)/(5a)`

And `(-y)/(-a^2 + 4ab) = 1/(-5ab)`

`=> -y = (-a^2 + 4ab)/(-5ab)`

`=> -y (a - 4b)/(5b)`

`=> y = (4b -a)/(5b)`

Hence `x = (4a -b)/(5a), y = (4b - a)/(5b)` is the solution of the given system of equation.

Page 8

The given system of equation is

5ax + 6by = 28

`=> 5ax + 6by - 28 = 0` .....(i)

and 3ax + 4by - 18 = 0

=> 3ax + 4by - 18 = 0 ....(ii)

Here

`a_1 = 5a , b_1 = 6b, c_1 = -28`

`a_2 = 3a,  b_2 = 4b, c_2 = -18`

By cross multiplication we have

`= x/(6b xx (-18) - (-28) xx 4b) = (-4)/(5a xx (-18) - (-28) xx 3a) = 1/(5a xx 4b - 6b xx 3a)`

`=> x/(-108b + 112b) = (-y)/(-90a + 80a) = 1/(20ab - 18ab)`

`=> x/(4b) = (-y)/(-6a) = 1/(2ab)`

Now

`x/(4b) = (-y)/(-6a) = 1/(2ab)`

Now

`x/(4b) = 1/(2ab)`

`=> x = (5b - 2a)/10ab`

And

`(-y)/(-6a) = 1/(2ab)`

`=> y = (6a)/(2ab) = 3/b`

Hence  `x = 2/a, y = 3/b` is the soluytion of the given system of equation.

Page 9

The given system of equations may be written as

(a + 2b)x + (2a − b)y - 2 = 0

(a − 2b)x + (2a + b)y - 3 = 0

Here,

`a_1 = a + 2b, b_1  = 2a - b, c_1 = -2`

`a_2 = a - 2b, b_2 = 2a + b, c_2 = -3`

By cross multiplication, we have

`=> x/(-3(2a - b) - (-2)(2a + b)) = (-y)/(3(a + 2b) - (-2)(a - 2b)) = 1/((a + 2b)(2a + b)-(2a - b)(a - 2b))`

`=> x/(-6a + 3b +4a + 2b) =  (-y)/(-3a - 6b + 2a - 4b) = 1/(2a^2 + ab + 4ab + 2b^2 -(2a^2 - 4ab - ab + 2b^2))`

`=> x/(-2a + 5b) = (-y)/(-a - 10b) = 1/(2a^2 + ab+ 4ab + 2b^2 - (2a^2 - 4ab - ab + 2b^2))`

`=> x/(-2a + 5b) = (-y)/(-a + 10b) = 1/(10ab)`

`=> x/(-2a + 5b) = y/(a+10b) = 1/(10ab)`

Now

`x/(-2a + 5b) = 1/(10ab)`

`=> y = (a + 10b)/(10ab)`

And

`y/(a + 10b) = 1/(0ab)`

`=> y = (a + 10b)/(10ab)`

Hence  `x = (5b - 2a)/(10ab), y = (a + 10b)/(10ab)` is the solution of the given system of equations

Page 10

The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

`x(a - b + (ab)/(a -  b)) = y(a + b - (ab)/(a + b)) = 0`

`x + y = 2a^2 = 0`

By cross multiplication method we get

`x/((-2a^2)xx-((a + b)- (ab)/(a + b)) - 0)  = (-y)/((-2a^2)xx((a - b) + (ab)/(a - b)) = 0)`

`= 1/((a - b)+ (ab)/(a - b) - (-((a + b) - (ab)/(a + b))))`

`x/((-2a^2)xx-((a+ b)^2 + ab)/(a + b)) = (-y)/((-2a^2)xx(((a - b)^2 + ab)/((a - b)))`

`= 1/((((a - b)^2 + ab)/(a - b)) - (-(((a + b)^2 - ab)/(a + b))`

`x/((-2a^2)xx-((a^2 + b^2 + 2ab) - ab)/(a + b)) = (-y)/((-2a^2)xx((a^2 + b^2 - 2ab) + ab)/(a - b))`

`= 1/((((a^2 + b^2 - 2ab) + ab)/(a - b)) - (-(((a^2 + b^2 + 2ab) - ab)/(a + b))`

`x/(((2a^4 + 2a^2b^2 + 2a^3b))/(a + b))= y/((2a^4 + 2a^2b^2 - 2a^3b)/(a - b))`

`= 1/(((a^2 + b^2 -ab )(a + b) + (a^2 + b^2 + ab)(a - b))/((a - b)(a + b)))`

`x/((2a^4 + 2a^2b^2 + 2a^3b)/(a + b)) = y/((2a^4 + 2a^2b^2 - 2a^3b)/(a - b)) = 1/(((2a^3)/((a - b)(a + b)))`

Consider the following

`x/((2a^4 + 2a^2b^2 + 2a^3b)/(a + b)) = 1/(((2a^3)/((a - b)(a + b))))`

`x = ((a^2 + b^2 + ab)(a - b))/a`

`x = ((a^2 + b^2 + ab)(a - b))/a`

`x = (a^3 + ab^2 + a^2b - b^3 -ab^2 - a^2b)/a`

`x = (a^3 - b^3)/a`

And

`y/((2^4 + 2a^2b^2 -2a^3b)/(a - b))  =1 /((2a^3)/((a  -b)(a + b)))`

`y/((a^2 + b^2 - ab)/(a - b)) = 1/(a/((a - b)(a + b)))`

`y(a/((a - b)(a + b))) = (a^2 + b^2 - ab)/(a - b)`

`y = ((a^2 + b^2 - ab)(a + b))/a`

`y = (a^3 + b^3)/a`

Hence we get the value of `x = (a^3 - b^3)/a and y = (a^3 + b^3)/a`

Page 11

The given system of equation is

bx + cy  = a + b  .....(i)

`ax (1/(a - b) - 1/(a + b)) + cy(1/(b -a) - 1/(b + a)) = (2a)/(a + b)`  ......(ii)

From equation (ii), we get

bx + cy - (a + b) = 0 .....(iii)

From equation (ii), we get

`ax[(a + b-(a-b))/((a - b)(a + b))] + cy((b + a - (b -a))/((b-a)(b + a))) - (2a)/(a + b) = 0`

`=> ax [(a + b - a + b)/((a - b)(a + b))] + cy((b + a - b + a)/((b -a)(b+a))) - (2a)/(a + b) = 0`

`=> ax[(2b)/(a - b)(a + b)] + cy((2a)/((b -a)(b+a))) - (2a)/(a + b) = 0`

`=> x [(2ab)/((a - b)(a + b))] + y((2ac)/((a - b)(a + b))) - (2a)/(a +b) = 0`

`=> 1/(a + b)[(2abx)/(a - b) - (2acy)/(a -b) - 2a] = 0`

`=> (2abx - 2acy - 2a(a - b))/(a - b) = 0`

`=> 2abx -  2acy - 2a(a - b) = 0` .....(iv)

From equation (i) and equation (ii), we get

`a_1 = b, b_1 = c, c_1 = -(a + b)`

`a_2 = 2ab,b_2 = -2ac, c_2 = -2a(a - b)`

By cross multiplication, we get

`=> x/(-2ac(a - b)-[-(a + b)][-2ac]) = (-y)/(-2ab(a - b)-[-(a + b)][2ab]) = 1/(-2abc - 2abc))`

`=> x/(-2a^2c + 2abc - 2a^2c - 2abc) = (-y)/(-2a^2b + 2ab^2 + 2ab^2b - 2ab^2) = (-1)/(4abc)`

`=> x/(-4a^2c) = (-y)/(4ab^2) = (-1)/(4abc)`

Now

`x/(-4a^2c) = (-1)/(4abc)`

`=> x = (4a^2c)/(4abc) = a/b`

And

`(-y).(4ab^2) = (-1)/(4abc)`

`=> y = (4ab^2)/(4abc) = b/c`

Hence `x = a/b, y = b/c` isthe soluiton of the given system of the equations.

Page 12

The given system of equation is

`(a - b)x + (a + b)y = 2a^2 - 2b^2`  .....(1)

(a + b)(a + y) =  4ab ....(ii)

`From equation (i), we get

`(a - b)x + (a + b)y - 2a^2 - 2b^2 = 0`

`=> (a - b)x + (a - b) y - 2(a^2 - b^2 ) = 0` .....(iii)

From equation (ii), we get

`(a + b)x + (a + b)y - 4ab = 0` .....(iv)

Here

`a_1 = a - b, b_1 = a+ b, c_1 = -2(a^2 - b^2)`

`a_2 = a + b, b_2 = a+ b, c_2 = -4ab`

By cross multiplication, we get

`=> x/(-4ab(a+b)+2(a^2 - b^2)(a + b)) = (-y)/(-4ab(a - b) + 2(a^2 - b^2)(a +b)) = 1/((a - b)(a + b)- (a +b)(a + b))`

`=> x/(2(a +b)[-2ab + a^2 - b^2]) = (-y)/(-4ab(a - b)+2[(a -b)(a + b)](a + b)) = 1/((a + b)[(a -b) - (a + b)])`

`=> x/(2(a + b)(a^2 - b^2 - 2ab)) = (-y)/(2(a - b)[-2ab + (a + b)(a+b)]) = 1/((a + b)[a - b - a - b])`

`=> x/(2(a + b)(a^2 - b^2 - 2ab)) = (-y)/(2(a - b)[-2ab + (a^2 + b^2 + 2ab)]) = 1/((a + b)(-2b)`

`=> x/(2(a + b)(a^2 - b^2 - 2ab)) = (-y)/(2(a -b)(a^2 + b^2)) = 1/(-2b(a + b))`

Now

`x/(2(a + b)(a^2 - b^2 - 2ab)) = 1/(-2b(a + b))`

`=> x = (a^2 - b^2 - 2ab)/(-b)`

`=> x = (-a^2 + b^2 + 2ab)/b`

`= (2ab - a^2 + b^2)/b`

Now

`(-y)/(2(a - b)(a^2 + b^2)) = 1/(-2ab(a + b))``

`=> y = ((a -b)(a^2 + b^2))/(b(a + b))`

Hence `x = (2ab - a^2 + b^2)/b, y = ((a - b)(a^2 + b^2))/(b(a + b))` is the solution of the given system of equation

`(-y)/(-a^2d^2 + b^2c^2) = 1/(a^4 - b^4)`

`=> -y = (-a^2d^2 + b^2c^2)/(a^4 - b^4)`

`=> y = (a^2d^2 - b^2c^2)/(a^4 - b^4)`

Page 13

The given system of equations may be written as

`a^2x + b^2y - c^2 = 0`

`b^2x + a^2y - d^2 = 0`

Here,

`a_1 = a^2, b_1 = b^2, c_1 = -c^2`

`a_2 = b^2, b_2 = a^2, c_2 = -d^2`

By cross multiplication, we have

`=> x/(-b^2d^2 + a^2c^2) = (-y)/(-a^2d^2 + b^2c^2) = 1/(a^4 - b^4)`

Now

`x/(-b^2d^2 + a^2c^2) = 1/(a^4 - b^4)`

`=> x = (a^2c^2 - b^2d^2)/(a^4 - b^4)`

And

`(-y)/(-a^2d^2 + b^2c^2) = 1/(a^4 - b^4)`

`=> -y = (-a^2d^2 + b^2c^2)/(a^4 - b^4)`

`=> y = (a^2d^2 - b^2c^2)/(a^4 - b^4)`

Hence `x = (a^2c^2 - b^2d^2)/(a^4 - b^4), y = (a^2d^2 - b^2c^2)/(a^4-b^4)` is the solution of the given system of the equations.

Page 14

Let `1/(x +y) = u and 1/(x -y ) = v` Then the given system of equations become

`57u + 6v = 5 => 57u + 6v - 5 = 0`

``38u + 21v = 9 => 38u + 21v - 9 = 0`

Here

`a_1 = 57, b_1 = 6, c_1 = -5`

`a_2 = 38, b_2 = 21, c_2 = -9`

By cross multiplication, we have

`=> u/(-54 + 105) = (-v)/(-513 + 190) = 1/(1193 - 228)`

`=> u/51 = (-v)/(-323) = 1/969`

`=> u/51 = v/323 = 1/969`

Now

`u/51= 1/969`

`=> u = 51/969`

`=> u = 1/19`

And

`v/(323) = 1/969`

`=> v = 323/969`

`=> v = 1/3`

Now

`u = 1/(x + y)``

`=> 1/(x + y)  =1/ 19`

`=>x + y = 19` ....(i)

And

`v = 1/(x - y)` 

`=> 1/(x -y) = 1/3`

=> x - y = 3 ...(ii)

Now adding eq i and ii

we get x = 11

And after substituting te value x in eq (ii)

we get y = 8

Hence  the value oof x = 11 and y = 8

Page 15

The given system of equation may be written as

2(ax – by) + a + 4b = 0

2(bx + ay) + b – 4a = 0

here

`a_1 = 2a, b_1 = -2b, c_1 = a+ 4b`

`a_2 = 2b, b_2 = 2a, c_2 = b - 4a`

By cross multiplication, we have

`=> x/(((-2b)(b - 4a) - (2a)(a + 4b))) = (-y)/((2b)(b - 4a) - (2a) (a + 4a) ) = 1/(4a^2 + 4b^2)`

`=> x/(-2b^2 + 8ab - 2a^2 - 8ab) = (-y)/(2ab - 8a^2 - 2ab - 8b^2) = 1/(4a^2 + 4b^2)`

`=> x/(-2a^2 - 2b^2) = (-y)/(-8a^2 - 8b^2) = 1/(4a^2 + 4b^2)`

Now

`x/(-2a^2 - 2b^2) = 1/(4a^2 + 4b^2)`

`=> x = (-2a^2 - 2b^2)/(4a^2 + 4b^2)`

`= (-2(a^2 - b^2))/(4(a^2 + b^2))`

`= (-1)/2`

And

`(-y)/(-8a^2 - 8b^2) = 1/(4a^2 + 4b^2)`

`=> -y = (-8a^2 - 8b^2)/(4a^2 + 4b^2)`

`=> -y = (-8(a^2 - b^2))/(4(a^2 + b^2))`

`=> -y = (-8)/4`

=> y = 2

Hence x = (-1)/2, y = 2 is the solution of the given system of the equations.

Page 16

The given system of equation is

6(ax + by) = 3a + 2b  .....(i)

6(bx - ay) = 3b - 2a .....(ii)

From equation (i), we get

6ax + 6by - (3a + 2b) = 0 .....(iii)

From equation (ii), we get

6bx - 6ay - (3b - 2a) = 0  .....(iv)

Here

`a_1 = 6a, b_1 = 6b,c_1 = -(3a + 2b)`

`a_2 = 6b, b_2 = -6a, c_2 = -(3b - 2a)`

By cross multiplication, we have

`x/(-6b(3b - 2a) - 6a (3a + 2b)) = (-y)/(-6a(3b - 2a) + 6b(3a + 2b)) = 1/(-36a^2 - 36b^2)`

`=> x/(-18b^2 + 12ab - 18a^2 - 12ab) = (-y)/(-18an + 12a^2 + 18ab + 12b^2) = 1/(-36a^2 - 36b^2)`

`=> x/(-18a^2 - 18b^2) = (-y)/(12a^2 + 12b^2) = 1/(-36(a^2 + b^2))`

`=> x/(-18(a^2 + b^2)) = (-y)/(12(a^2 + b^2)) = (-1)/(36(a^2 + b^2))`

Now

`x/(-18(a^2 + b^2)) = (-1)/(36(a^2 + b^2))`

`=> x = (18(a^2 + b))/(36(a^2 + b^2))`

= 1/2

And

`(-y)/(12(a^2+b^2)) = (-1)/(36(a^2 + b^2))`

`=> y = (12(a^2 + b^2))/(36(a^2 + b^2))`

y = 1/3

Hence x = 1/2, y = 1/3 is the solution of the given system of equations. 

Page 17

taking `1/x = u and 1/y = v` Then the given system of equations become

`a^2u - b^2v = 0`

`a^2bu + b^2av - (a + b) = 0`

Here

`a_1 = a^2, b_1 = -b^2, c_1 = 0`

`a_2 = a^2b, b_2 = b^2a, c_2 = -(a + b)`

By cross multiplication, we have

`=> u/(b^2(a + b)-0xxb^2a) = (-v)/(-a^2(a + b)-0xxa^2b) = 1/(a^3b^2 + a^2b^3)`

`=> u/(b^2(a + b)) = v/(a^2(a + b)) = 1/(a^2b^2(a + b))`

Now

`u/(b^2(a + b)) = 1/(a^2b^2(a + b))`

`=> u = (b^2(a + b))/(a^2b^2 (a + b))`

`=> u = 1/a^2

And

`v/(a^2(a + b)) = 1/(a^2b^2(a + b))`

`=> v = (a^2 (a + b))/(a^2b^2 (a + b))`

`=> v = 1/b^2`

Now

`x = 1/u = a^2`

And

`y = 1/v = b^2`

Hence `x = a^2, y = b^2` is the solution of the given system of equations

Page 18

The given system of equations may be written as

`mx - ny - (m^2 + n^2) = 0`

`x + y - 2m = 0

Here

`a_1 = m, b_1 = -n, c_1 = -(n^2 + n^2)`

`a_2 = 1, b_2 = 1, c_2 = -2m`

By cross multiplication, we have

`x/(2mn + (m^2 + n^2)) = (-y)/(-2m^2 + (m^2 + n^2)) = 1/(m + n)`

`=> x/(2mn + m^2 + n^2) = (-y)/(-m^2 + n^2) = 1/(m + n)`

`=> x/(m + n)^2 = (-y)/(-m^2 + n^2) = 1/(m + n)`

Now

`x/(m + n)^2 = 1/(m + n)`

`=> x = (m + n^2)/(m + n)`

=> x = m + n

And

`(-y)/(-m^2 + n^2)  = 1/(m + n)`

`=> -y = (-m^2 + n^2)/(n + n)`

`=> y = (m^2 - n^2)/(m + n)`

`=> y = ((m - n)(m + n))/(m + n)`

=> y = m - n

Hence, x = m + n, y = m - n is the solution of the given system of equation.

Page 19

The given system of the equation may be written as

`a/b x xx - b/a xx y - (a + b) = 0`

ax - by - 2ab = 0

Here

`a_1 = a/b, b_1 = - b/a, c_1 = -(a + b)`

`=> x/(2b^2 - ab - b^2) = (-y)/(-2a^2 + a^2 + ab) = 1/(-a + b)`

`=>  x/(b^2 -ab) = (-y)/(-a^2 + ab) = 1/(-a + b)`               

`=> x/(b(b -a)) = (-y)/(a(-a + b)) = 1/(b - a)` 

Now

`x/(b(b - a))  = 1/(b -a)`                   

`=> x = (b(b -a ))/(b -a) = b`

And

`=> -y = (a(b - a))/(b -a)`

=> -y = a

=> y = -a

Hence, x = b, y = -a is the solution of the given system of equations.

Page 20

The given system of the equation may be written as

`b/a x + a/b y - (a^2 + b^2) = 0``

x + y - 2ab = 0

Here

`a_1 = b/a, b_1 =- a/b,c_1 = -(a^2 + b^2)`

`a_2 = 1, b_2 = 1, c_2 = -2ab`

By cross multiplication, we have

`x/(-2ab  xx a/b + a^2 + b^2) = (-y)/(-2ab xx a/b + a^2 + b^2) = 1/(b/a - a/b)`

`=> x/(-2a^2 + a^2 + b^2) = (-y)/(-2b^2 + a^2 + b^2) = 1/((b^2 - a^2)/(ab))`

`=> x/(b^2 - a^2) = (-y)/(-b^2 + a^2) = 1/((b^2 - a^2)/(ab))`

Now

`x/(b^2 - a^2) = 1/((b^2 - a^2)/(ab))`

`=> x = b^2 - a^2 aa (ab)/(b^2 - a^2)`

=> x = ab

And

`(-y)/(-b^2 + a^2) = 1/((b^2 - a^2)/(ab))`

`=> -y = b^2 + a^2 xx (ab)/(b^2 - a^2)`

`=> -y =-(b^2 - a^2) xx (ab)/(b^2 - a^2)``

=> -y = -ab

=> y = ab

Hence, x = ab,y = ab is the solution of the given system of equations.

Page 21

The given system of the equation may be written as

`1/a x xx + 1/b xx y -(a + b) = 0`

`1/a^2 x xx 1/b^2 xx y - 2 = 0`

Here

`a_1 = 1/a, b_1 = 1/b, c_1 = -(a+ b)`

`a_2 = 1/a^2, b_2 = 1/b^2 , c_2 = -2`

By cross multiplication, we ge

`=> x/(1/b xx (-2) - 1/(b^2) x - (a +b)) = (-y)/(1/\a xx -2 - 1/a^2 x - (a + b)) = 1/(1/a xx 1/b^2 - 1/a^2 xx 1/b)`

`=> x/(-2/b + a/b^2 + 1/b) = (-y)/(-2/a + 1/a + b/a^2) = 1/(-1/(ab^2) - 1/(a^2b))`

`=> x/(a/b^2 - 1/b) = (-y)/(-1/a + b/a^2) = 1/(1/(ab^2) - 1/(a^2b))`

`=> x/((a- b)/b^2) = y/((a-b)/a^2) = 1/((a - b)/(a^2b^2))`

`=> x = (a - b)/b^2 xx 1/((a - b)/(a^2b^2)) = a^2` and

`y = (a - b)/a^2 xx 1/((a - b)/(a^2b^2)) = b^2`

Hence `x = a^2, y= b^2` is solution of the given system of the equtaions.