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Solve the following pair of linear equations: 10/(x + y) − 1/(x − y) = 1, 5/(x + y) + 3/(x − y) = 4 - Mathematics

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Question

Solve the following pair of linear equations:

`10/(x + y) - 1/(x - y) = 1, 5/(x + y) + 3/(x - y) = 4`

Sum
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Solution

Given:

`10/(x + y) - 1/(x - y) = 1`

`5/(x + y) + 3/(x - y) = 4`

Step 1: Let `u = 1/(x + y), v = 1/(x - y)`

Rewrite the given equations in terms of (u) and (v):

10u – v = 1 

5u + 3v = 4

Step 2: Solve the system for (u) and (v).

Multiply the second equation by 3 to align the coefficients of (v):

15u + 9v = 12

Multiply the first equation by 3:

30u – 3v = 3

Add the two scaled equations:

(30u – 3v) + (15u + 9v) = 3 + 12

⇒ 45u + 6v = 15

This doesn’t isolate well; instead, use elimination for (v) from original system:

Multiply the first equation by 3:

30u – 3v = 3

Add to the second equation:

5u + 3v = 4

Add:

(30u – 3v) + (5u + 3v) = 3 + 4 

⇒ 35u = 7

⇒ `u = 7/35`

⇒ `u = 1/5`

Step 3: Substitute `(u = 1/5)` in first equation:

`10 xx 1/5 - v = 1`

⇒ 2 – v = 1

⇒ v = 1

Step 4: Find (x + y) and (x – y): 

`u = 1/(x + y)`

`u = 1/5`

⇒ x + y = 5

`v = 1/(x - y)`

v = 1

⇒ x – y = 1

Step 5: Solve for (x) and (y):

Add the two equations:

x + y = 5 

x – y = 1

2x = 6

⇒ x = 3

Substitute (x = 3) into (x + y = 5):

3 + y = 5 

⇒ y = 2

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Chapter 5: Simultaneous Linear Equations - Exercise 5D [Page 110]

APPEARS IN

Nootan Mathematics [English] Class 9 ICSE
Chapter 5 Simultaneous Linear Equations
Exercise 5D | Q 11. | Page 110
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