Solve the following pair of linear equations by the elimination method and the substitution method: x + y =5 and 2x –3y = 4 3x + 4y = 10 and 2x – 2y = 2 3x – 5y – 4 = 0 and 9x = 2y + 7 x/2 + 2y /3 = - 1 and x – y/3 = 3

Solve the following pair of linear equations by the elimination method and the substitution method:

x + y =5 and 2x –3y = 4

3x + 4y = 10 and 2x – 2y = 2

3x – 5y – 4 = 0 and 9x = 2y + 7

 x/2 + 2y /3 = - 1 and x – y/3 = 3 

Solution 

x + y =5 and 2x –3y = 4

By elimination method

x + y =5          ……….(1)

2x –3y = 4      ………(2)

Multiplying equation (1) by 2, we obtain

2x + 2y = 10   ………..(3)

2x –3y = 4      ………(2)

Subtracting equation (2) from equation (3), we obtain

5y = 6

Y = 6/5

Substituting the value in equation (1), we obtain

X = 5  - (6/5)   = 19/5

 So our answer is x = 19/5 and y = 6/5

By substitution method

x + y =5          ……….(1)

subtract y both side we get

x = 5 - y          ……..,(4)

plug the value of x in equation second we get

2(5 – y ) – 3y  = 4

-5y = - 6

Y = -6/-5 = 6/5

Plug the value of y in equation 4 we get

X         = 5 – 6/5

X         = 19/5

So our answer is x = 19/5 and y = 6/5 again

3x + 4y = 10 and 2x – 2y = 2

 By elimination method

3x + 4y = 10    ………(1)

 2x – 2y = 2    ……….(2)

Multiplying equation (2) by 2, we obtain

4 x – 4 y  = 4 ………..(3)

3x + 4y = 10    ………(1)

Adding equation (1) and (3), we obtain

7x  +  0  = 14

Divide by 7 both side we get

X = 14/7 = 2

Substituting in equation (1), we obtain

3x + 4y = 10

3(2) +  4 y = 10

6 +  4 y  = 10

4y        = 10 – 6

4Y       = 4

Y         = 4/4 = 1

Hence answer is  x = 2, y = 1

By substitution method

3x + 4y = 10                           ………(1)

Subtract 3x both  side we get

4 y = 10 – 3x

Divide by 4 we get

Y = (10 - 3x )/4

Plug this value in equation second we get

 2x – 2y = 2                            ……….(2)

2x – 2(10 - 3x )/4) = 2

 Multiply by 4 we get

8x  - 2(10 – 3x) = 8

8x  - 20 + 6 x = 8

14 x  = 28

 X = 28/14 = 2

Y = (10 - 3x )/4

Y = 4/4 = 1

3x – 5y – 4 = 0 and 9x = 2y + 7

 By elimination method

3x – 5y – 4 = 0

3x – 5y           = 4  ………….(1)

9x = 2y + 7

9x – 2y  = 7    ……….(2)

Multiplying equation (1) by 3, we obtain

9 x – 15 y = 11          ……(3)

9x – 2y  = 7                ...….(2)

Subtracting equation (2) from equation (3), we obtain

-13 y  = 5

Y = - 5/13

Substituting in equation (1), we obtain

3x – 5y           = 4  ………….(1)

3x  - 5(-5/13)  = 4

Multiply by 13 we get

39 x  + 25  = 52

39 x     = 27

            X =27/39 = 9 /13

Hence our answer Is x = 9/13 and y = - 5/13

By substitution method

3x – 5y           = 4  ………….(1)

Add 5y both side we get

3x = 4 + 5y

Divide by 3 we get

X = (4 + 5y )/3            ……..(4)

Plug this value in equation second we get

9x – 2y  = 7                ...….(2)

9 ((4 + 5y )/3) – 2y = 7

Solve it we get

3(4 + 5y ) – 2y = 7

12 + 15 y – 2y = 7

13 y = - 5

Y = -5/13

Plug this value back in equation 4 we get

X = (4 + 5y )/3

X = (4 +5* (-5/13))/ 3                  

                        Hence we get x = 9/13 and y = - 5/13 again

x/2 + 2y /3 = - 1 and x – y/3 = 3

(iv)      By elimination method

x/2 + 2y /3 = - 1         ………..(1)

x – y/3 = 3                  ………..(2)

Multiplying equation (1) by 2, we obtain

x + 4y/3 = - 2 ………(3)

x – y/3 = 3                  ………..(2)

Subtracting equation (2) from equation (3), we obtain

5y /3  = -5

Divide by 5 and multiply by 3 we get

Y = -15/5

Y         = - 3

Substituting in equation (2), we obtain

x – y/3 = 3                  ………..(2)

x – (-3)/3 = 3

x +  1 = 3

x = 2

Hence our answer is  x = 2 and y = −3

By substitution method

x – y/3 = 3                  ………..(2)

Add y/3 both side we get

x= 3 + y/3       ……(4)

                                    Plug this value in equation (1) we get

x/2 + 2y /3 = - 1         ………..(1)

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

3/2 +  y /6 + 2y/3 = - 1

Multiply by 6 we get

9 +  y + 4y  = - 6

5y = -15

Y = - 3

Hence our answer is  x = 2 and y = −3