Question

Solve the following systems of equations:
 x/2 + y = 0.8
7/(x+ y/2) = 10

Answer 1

The given system of equations can be solved by putting x as 2 / 5, and y as 3 / 5.

• The first equation in the system is given as :

(x / 2) + y = 0.8

=> (x + 2y) / 2 = 0.8

=> (x + 2y) = 0.8 × 2

=> x + 2y = 1.6

=> x + 2y = 16 / 10

=> x + 2y = 8 / 5

=> 5 (x + 2y) = 8

=> 5x + 5.2y = 8

=> 5x + 10y = 8 -(i)

• The second equation in the system is given as :

7 / (x + y / 2) = 10

Following the B.O.D.M.A.S. rule,

7 / { (2x + y) / 2 } = 10

=> (7 × 2) / (2x + y) = 10

=> 14 / (2x + y) = 10

=> 14 = 10 (2x + y)

=> 14 = 10.2x + 10y

=> 14 = 20x + 10y

=> 14 = 2 (10x + 5y)

=> 14 / 2 = 10x + 5y

=> 7 = 10x + 5y

=> 10x + 5y = 7 -(ii)

• Multiplying eq. (ii) by 2, we get,

2 ( 10x + 5y) = 2 × 7

=> 20x + 10y = 14   -(iii)

• Subtracting eq. (iii) from (i), we get,

(i) - (iii)

=> (5x - 20x) + (10y - 10y) = 8 - 14

=> - 15x + 0 = - 6

=> - 15x = - 6

=> x = - 6 / - 15

=> x = 2 / 5

• Putting x = 2 / 5 in eq. (i), we get the value of y.

∴  5 × (2 / 5) + 10y = 8

=> (5 × 2) / 5 + 10y = 8

=> 2 + 10y = 8

=> 10y = 8 - 2

=> 10y = 6

=> y = 6 / 10

=> y = 3 / 5

• ∴  The values are :

x = 2 / 5

y = 3 / 5