4x 3y = 14 3x 4y = 23 by elimination method

Solve the following systems of equations:

`4/x + 3y = 14`

`3/x - 4y = 23`

`4/x + 3y = 14`

`3/x - 4y = 23`

Let `1/x = p`

The given equations reduce to:

4p + 3y = 14 

=> 4p + 3y - 14 = 0 ...(1)

3p - 4y = 23

=> 3p - 4y - 23 = 0 ....(2)

Using cross-multiplication method, we obtain

`p/(-69-56) = y/(-42 - (-92)) = 1/(-16 - 9)`

`p/(-125) = y/50 = (-1)/25`

p = 5 , y = -2

`:. p = 1/x = 5`

`x = 1/5`

Concept: Algebraic Methods of Solving a Pair of Linear Equations - Cross - Multiplication Method

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Page 2

Solve the following systems of equations:

`x+y = 2xy`

`(x - y)/(xy) = 6`   x != 0, y != 0

The system of the given equation is

x + y = 2xy .....(i)

And `(x - y)/(xy) = 6`

x - y = 6xy .....(ii)

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

2x = 2xy + 6xy

=> 2x = 8xy

`=> (2x)/(8x) = y`

`=> y = 1/4`

Putting y = 1/4 in equation (i) we get

`x + 1/4 = 2x xx 1/4`

`=> x + 1/4 = x/2`

`=> x - x/2 = (-1)/4`

`=> (2x - x)/2 = (-1)/4`

`=> x = (-2)/4 = (-1)/2`

Hence, solution of the given system of equation is `x = (-1)/2, y = 1/4`

Concept: Algebraic Methods of Solving a Pair of Linear Equations - Substitution Method

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Page 3

The system of the given equation is

2(3u − ν) = 5uν

=> 6u - 2v = 5uv ....(i)

And 2(u + 3v) = 5uv

=> 2u + 6v = 5uv .....(ii)

Multiplying equation (i) by 3 and equation (ii) by 1, we get

18u - 6v = 15uv ...(iii)

2u + 6v = 5uv ...(iv)

Adding equation (iii) and equation (iv), we get

18u + 2u = 156uv + 5uv

=> 20u = 20uv

`=> (20u)/(20u)= v`

=> v = 1

Putting v = 1in equation (i), we get

6u - 2 x 1 = 5u x 1

=> 6u - 2 = 5u

=> 6u - 5u = 2

=> u = 2

Hence, solution of the given system of equation is u = 2, v=- 1

Page 4

Let `1/(3x + 2y) = u and 1/(3x - 2y) = v` Then, the given system of equation becomes

`2u + 3v = 17/5` ....(i)

5u + v = 2 ....(ii)

Multiplying equation (ii) by 3, we get

`15u - 2u = 6 - 17/5`

`=> 13u = (30 - 17)/5`

`=> 13u = 13/5`

`=> u = 13/(5 xx 13) = 1/5`

Putting u = 1/5 in eqaution  (ii) we get

`5 xx 1/5 + v = 2`

=> 1 + v = 2

=> v = 2 - 1

=> v = 1

Now `u = 1/(3x + 2y)`

`=> 1/(3x + 2y) = 1/5`

=> 3x + 2y = 5 ....(iv)

And `v = 1/(3x + 2y)`

`=> 1/(3x + 2y ) = 1/5`

=> 3x + 2y = 5 ....(iv)

And `v = 1/(3x + 2y)`

=> 3x - 2y = 1 ...(v)

Adding equation (iv) and (v), we get

6x = 1 + 5

6x = 6

x = 1

Putting 1 x  in equation (v), we get

3 xx 1 + 2y = 5

2y = 5 - 3

2y  = 2

y = 2/2 = 1

Hence, solution of the given system of equation is x = 1, y = 1

Page 5

We have,

x − y + z = 4  ...(i)

x + y + z = 2 ....(ii)

2x + y − 3z = 0 ....(iii)

From equation (i), we get

z = 4 - x + y

z = -x + y + 4

Substituting z = -x + y + 4 in equation (ii), we get

x + y + (-x + y + 4) = 2

=> x + y - x + y + 4 = 2

=> 2y + 4 = 2

`=> 2y = 2 - 4 = -2`

=> 2y = -2

`=> y = (-2)/2 = -1`

Substituting the value of z in equation (iii), we get

2x + y -3(-x + y + 4) = 0

=> 2x + y + 3x - 3y - 12 = 0

=> 5x - 2y - 12 = 0

=> 5x - 2y  = 12 ....(iv)

Putting y = -1 in equation (iv), we get

`5x - 2xx (-1) = 12`

=> 5x + 2 = 12

=> 5x = 12 - 2 = 10

`=> x = 10/5 = 2`

Putting x = 2 and y = -1 in z = -x + y + 4 we get

z = -2 + (-1) + 4

=-2 - 1 + 4

= -3 + 4

= 1

Hence, solution of the giving system of equation is x = 2, y = -1, z = 1

Page 6

Let `1/(x + y) = u and 1/(x - y) = v`

Then, the system of the given equations becomes

44u + 30v = 10 ....(i)

55u + 40v = 13 ....(ii)

Multiplying equation (i) by 4 and equation (ii) by 3, we get

176u + 120v = 40 ...(iii)

165u + 120v  = 39 ...(iv)

Subtracting equation (iv) by equation (iii), we get

176 - 165u = 40 - 39

=> 11u = 1

`=> u = 1/11`

Putting u = 1/11 in equation (i) we get

`44 xx 1/11 + 30v = 10`

4 + 30v = 10

=> 30v = 10 - 4

=> 30v = 6

`=> v = 6/30 = 1/5`

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

`=> 1/(x + y) = 1/11`

=> x + y = 11 ...(v)

Adding equation (v) and (vi), we get

2x = 11 + 5

=> 2x = 16

`=> x = 16/2 = 8`

Putting x = 8 in equation (v) we get

8 + y = 11

=> y = 11 - 8 - 3

Hence, solution of the given system of equations is x = 8, y = 3

Page 7

`10/(x + y) + 2/(x - y)  = 4`

`15/(x + y) - 5/(x - y) = -2`

Let `1/(x + y) = p and 1/(x - y) = q`

The given equations reduce to:

10p + 2q = 4

=> 10p + 2q - 4 = 0 ....(1)

15p - 5q = -2 

=> 15p - 5q + 2 = 0 ...(2)

Using cross-multiplication method, we obtain:

`p.(4- 20) = q/(-60-20) = 1/(-50-30)`

`p/(-16) = q/(-80) = 1/(-80)`

p = 1/5 and q =1

p = 1/(x + y) = 1/5 and `q = 1/(x - y) = 1`

x + y = 5 .....(3)

x - y = 1 ....(4)

Adding equation (3) and (4), we obtain:

2x = 6

x = 3

Substituting the value of x in equation (3), we obtain:

y = 2

∴ x = 3, y = 2

Page 8

The given system of equation is

`4/x + 15y = 21` ....(i)

`3/x + 4y = 5` ....(ii)

Multiplying equation (i) by 3 and equation (ii) by 4, we get

`12/x + 15y = 21`   .....(iii)

`12/x + 16y = 20` .....(iv)

Subtracting equation (iii) from equation (iv), we get

`12/x - 12/x + 16y - 15y = 20 - 21`

y = -1

Putting y = -1 in equation (i) we get

`4/x + 5xx (-1) = 7`

`=> 4/x - 5 = 7`

`=> 4/x = 7 + 5`

`=> 4/x = 12`

=> 4 = 12x

=> 4/12 = x

`=> x = 4/12`

`=> x = 1/3`

Hence, solution of the given system of equation x = 1/3, y =- -1

Page 9

Let us write the given pair of equation as

`2(1/x) + 3(1/y) = 13`  ....(1)

`5(1/x) - 4(1/y) = -2` ....(2)

These equation are not in the form ax + by + c = 0 However, if we substitute

`1/x = p` and `1/y = q` in equation (1) and (2) we get

2p + 3q = 13

5p - 4q = -2

So, we have expressed the equations as a pair of linear equations. Now, you can use any method to solve these equations, and get p = 2, q = 3

You know that `p =1/x and q =1/y`

Substitute the values of p and q to get

`1/x = 2 i.e x = 1/2 and 1/y = 3 " i.e " y = 1/3`

Page 10

The given equation is

`1/(3x + y) + 1/(3x - y) = 3/4`

`1/(2(3x + y)) - 1/(2(3x - y)) = -1/8`

Let `1/(3x + y) = u and 1/(3x - y) = v` then equation are

`u + v = 3/4` ...(i)

`u/2 - v/2 = 1/8` ....(ii)

Multiply equation (ii) by 2 and add both equations, we get

`u + v = 3/4`

`u - v = -1/4`

`2u = 1/2`

`u = 1/4`

Put the value of u  in equation (i) we get

`1 xx 1/4 + v = 3/4`

`v= 1/2`

Then

`1/(3x + y) = 1/4`   ....(iii)

3x  + y = 4

`1/(3x -y) = 1/2`  ...(iv)

3x - y = 2

Add both equation we get

3x + y = 4

3x - y = 2

_________

6x = 6

x = 1

Put the value of x in equation (iii) we get

3 x 1 + y = 4

y = 1

Hence value of x = 1 and  y = 1

Page 11

Solve the following systems of equations:

`5/(x - 1) + 1/(y - 2)  = 2`

Let us put` 1/(x  - 1) = p and 1/(y - 2) = q` The the given equation

`5(1/(x - 1)) + 1/(y - 2) = 2` .....(1)

`6(1/(x - 1))  -3 (1/(y - 2) ) = 1` ....(2)

Can be written as 

5p + q = 2 ....(3)

6p - 3q = 1 ...(4)

Equation 3 and 4 from a pair of linear equations in the genera; form. Now you can use any method to solve these equation we get p = 1/3 and q = 1/3 now

substituting `1/(x - 1)` fpr p we have

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

i.e x - 1 = 3 i.e x = 4

Similary substituting 1/(y -2) for q we get

`1/(y -2) = 1/3`

i.e 3 = y - 2 i.e y = 5

Hence x = 4, y = 5 is required solution of the given pair of euation

Concept: Algebraic Methods of Solving a Pair of Linear Equations - Substitution Method

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Page 12

Solve the following systems of equations:

`(7x - 2y)/"xy" = 5`

`(8x + 7y)/"xy" = 15`

`(7x - 2y)/"xy" = 5`

`=> 7/y - 2/x = 5` .....(1)

`(8x + 7y)/(xy) = 15`

`=> 8/y + 7/x = 15` ..........(2)

Let `1/x = p and 1/y = q`

The given equations reduce to:

-2p + 7q = 5

=> -2p + 7q - 5 = 0 ...(3)

7p + 8q = 15

=> 7p + 8q - 15 = 0 ....(4)

Using cross multiplication method, we obtain:

`p/(-105-(-40)) = q/(-35-30) = 1/(-16-49)`

`p/(-65) = 1/(-65), q/(-65) = 1/(-65)`

p = 1, q= 1

`p = 1/x = 1, q= 1/y =1`

x = 1 , y = 1

Concept: Algebraic Methods of Solving a Pair of Linear Equations - Cross - Multiplication Method

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Page 13

Solve the following systems of equations:

152x − 378y = −74
−378x + 152y = −604

152x − 378y = −74 ...(1)

−378x + 152y = −604 ...(2)

Adding the equations (1) and (2), we obtain:

-226x - 226uy = -678

=> x + y = 3 ...(3)

Subtracting the equation (2) from equation (1), we obtain

530x - 530y = 530

=> x - y = 1 ...(4)

Adding equations (3) and (4), we obtain:

2x = 4

x = 2

Substituting the value of x in equation (3), we obtain:

y = 1

Concept: Algebraic Methods of Solving a Pair of Linear Equations - Substitution Method

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Page 14

The given system of equation is

99x + 101y = 499   ...(i)

101x + 99y = 501 ...(ii)

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

99x + 101x + 101y + 99y = 499 + 501

=> 200x + 200y = 1000

`=> 200(x + y) = 1000`

`=> x + y = 1000/200 = 5`

=> x + y = 5 ....(iii)

Subtracting equation (i) by equation (ii), we get

101x - 99x + 99y - 101y = 501 - 499

=> 2x - 2y = 2

=> 2(x - y)  = 2

=> x - y = 2/2

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

Adding equation (iii) and equation (iv), we get

2x = 5 + 1

`=> x = 6/2 = 3`

Putting x = 3 in equation (iii), we get

3 + y = 5

=> y = 5 - 3 = 2

Hence, solution of the given system of equation is x = 3, y = 2

Page 15

The given system of equation is

23x − 29y = 98 ....(i)

29x − 23y = 110 ...(ii)

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

23x + 29x - 29y - 23y = 98 + 110

=> 52x - 52y = 208

=> 52 (x - y) = 208

`=> x - y = 208/52 = 4`

=> x - y - 4 ...(iii)

Subtracting equation (i) by equation (ii), we get

29x - 23x - 23y + 29y = 110 - 98

=> 6x + 6y = 12

=> 6(x + y) = 12

`=> x + y = 12/6 = 2`

=> x = y = 2 ...(iv)

Adding equation (iii) and equation (iv), we get

2x = 2 + 4 = 6

Putting x = 3 in equation (iv), we get

3 + y = 2

=>? y = 2- 3 = -1

Hence, solution of the given system of equation is x = 3 , y = -1

Page 16

We have,

x − y + z = 4  .....(i)

x − 2y − 2z = 9 ....(ii)

2x + y + 3z = 1 .....(iii)

From equation (i), we get

z = 4 - x + y

=> z = -x + y + 4

Subtracting the value of z in equation (ii), we get

x - 2y - 2(-x + y + 4) = 9

`=> x - 2y + 2x - 2y - 8 = 8`

=> 3x - 4x = 9 + 8

=> 3x - 4y = 17 ......(iv)

Subtracting the value of z in equation (iii), we get

2x + y + 3(-x + y + 4) = 1

=> 2x + y + 3x + 3y + 12 = 1

=> -x + 4y =1 - 12

=> -x + 4y = -11 .....(v)

Adding equations (iv) and (v), we get

3x - x - 4y + 4y = 17 - 11

=> 2x = 6

=> x = 6/2 = 3

Putting x = 3 in equation (iv), we get

`3 xx 2  - 4y = 17`

=> 9 - 4y = 17

=> -4y = 17 - 9

=> -4y = 8

`=> y = 8/(-4) = -2`

Putting x = 3 and y = -2 in z = -x + y + 4 we get

z = -3 -2 + 4

=> x = -5 + 4

=> z= -1

Hence, solution of the giving system of equation is x = 3, y = -2, z = -1