Question 1. Solve the Following Pair of Linear Equations By the Exercise 3.4 Chapter 3 Pair of Linear Equations in

Question 1. 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: 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

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