Solve simultaneous linear equations: (i) 20/ (x + 1) + 4/ (y – 1) = 5

by

Solve simultaneous linear equations:

(i) 20/ (x + 1) + 4/ (y – 1) = 5
10/ (x + 1) – 4/ (y – 1) = 1
(ii) 3/ (x + y) + 2/ (x – y) = 3
2/ (x + y) + 3/ (x – y) = 11/3.

Solution:

(i) 20/ (x + 1) + 4/ (y – 1) = 5 ….. (1)
10/ (x + 1) – 4/ (y – 1) = 1 ….. (2)
Add equation (1) and (2)
30/ (x + 1) = 6
By cross multiplication
30 = 6 (x + 1)
By further calculation
30/6 = x + 1
5 = x + 1
So we get
x = 5 – 1 = 4
Substitute the value of x in equation (1)
20/ (x + 1) + 4/ (y – 1) = 5
20/ (4 + 1) + 4/ (y – 1) = 5
By further calculation
20/5 + 4/ (y – 1) = 5
4 + 4/ (y – 1) = 5
We can write it as
4/ (y – 1) = 5 – 4 = 1
4/ (y – 1) = 1
By cross multiplication
4 = 1 (y – 1)
So we get
4 = y – 1
y = 4 + 1 = 5
Therefore, x = 4 and y = 5.
(ii) 3/ (x + y) + 2/ (x – y) = 3 …. (1)
2/ (x + y) + 3/ (x – y) = 11/3 ….. (2)
Multiply equation (1) by 3 and (2) by 2
9/ (x + y) + 6/ (x – y) = 9 ….. (3)
4/ (x + y) + 6/ (x – y) = 22/3 ….. (4)
Subtracting both the equations
5/ (x + y) = 9 – 22/3
Taking LCM
5/ (x + y) = 5/3
By cross multiplication
5 × 3 = 5 (x + y)
By further calculation
(5 × 3)/ 5 = x + y
x + y = (3 × 1)/ 3
x + y = 3 …… (5)
Substitute equation (5) in (1)
3/3 + 2/ (x – y) = 3
By further calculation
1 + 2/ (x – y) = 3
2/ (x – y) = 3 – 1 = 2
So we get
2/2 = x – y
Here
1 = x – y ….. (6)
We can write it as
x – y = 1
x + y = 3
By adding both the equations
2x = 4
x = 4/2 = 2
Substitute x = 2 in equation (5)
2 + y = 3
y = 3 – 2 = 1
Therefore, x = 2 and y = 1.

More Solutions:

Leave a Comment