Question

Solve the following system of equations by the method of substitution :

i) x + y = 7, 2x – 3y = 11
ii) 2x – 7y = 1, 4x + 3y = 15
iii) 3x – 5y = 1, 5x + 2y = 19
iv) 5x + 8y = 9, 2x + 3y = 4
v) 4x + 9y = 5, 3x – 5y = 39
vi) x – y = 26, x = 3y

Answer 1

1) y = 7 - x put this in equation
2x -3y = 11
2x -3(7-x) = 11
5x -21 = 11
x = 32/5 so, y = 7-32/5 = 3/5

2) y = (2x -1)/7 put in equation
4x +3y = 15
4x + 3(2x -1)/7 = 15
28x + 6x -3 == 105
x= 108/34 = 54/17
So, y = (2x -1)/7 = (108-17)/119
= 91/119

3) 3x -5y = 1
y = (3x -1)/5 put in equation
5x +2(3x -1)/5 = 19
25x +6x -2 = 95
31x = 97
x = 97/31
so, y = (3x-1)/5 = (291-31)/155
= 260/155
4) 5x + 8y = 9
2x +3y = 4
y = (4-2x)/3 put 1st equation
5x +8(4-2x)/3 = 9
15x +32 -16x = 27
-2x = -5
x = 5/2 so, y = (4-2x)/3 = (4-5)/3 = -1/3

5) 4x +9y = 5 and 3x -5y =39
y =(5-4x)/9 put 2nd equation

3x -5(5-4x)/9 = 39
27x -25 + 20x = 351
47x = 376
x = 376/47 = 8
So, y = (5-4×8)/9 = -27/9 = -3

6) x - y = 26 and x = 3y
x =3y put in 1st equation ,
3y -y = 26 , y = 13
So, x = 3×13 = 39

Answer 2

second answer this questions