#### 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.

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