ALGEBRA J3

 

Solve the following equations by substitution method.

x + y = 3

x + 3y = -3

 

Solution

 

x + y = 3 ----------- (i)

x + 3y = -3 ---------- (ii)

 

from equation (i)

 

x + y = 3

x = 3 - y ----------(iii)

 

substitute equation (iii) in (ii) above,

 

x + 3y = -3

 

(3 - y) + 3y = -3

3 - y + 3y = -3  

3 + 2y = -3  

2y = -3 - 3 

 

2y = -6           [since –y+3y=2y]

 

 

 

2y = -6

2      2

 

y = -3

 

 

From equation (iii)

x = 3 - y

x = 3 – (-3) ----------(iii)

x = 3+3

x=6

Hence x=6 and y=-3.

 

 

TRY THIS………………………..

 

 

5x - 2y = 14

x  -  y = 4