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Tamil Nadu Board of Secondary EducationSSLC (English Medium) Class 10

Discuss the nature of solutions of the following system of equations y+z4=z+x3=x+y2; x + y + z = 27 - Mathematics

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Question

Discuss the nature of solutions of the following system of equations

`(y + z)/4 = (z + x)/3 = (x + y)/2`; x + y + z = 27

Sum

Solution

`(y + z)/4 = (z + x)/3`

3y + 3z = 4z + 4x

– 4x + 3y + 3z – 4z = 0

– 4x + 3y – z = 0

4x – 3y + z = 0   ...(1)

`(z + x)/3 = (x + y)/2`

2z + 2x = 3x + 3y

– 3x + 2x – 3y + 2z = 0

– x – 3y + 2z = 0

x + 3y – 2z = 0   ...(2)

                   x + y + z = 27     ....(3)
                 4x – 3y + z = 0      ....(1)
                 x + 3y – 2z = 0      ....(2)
(1) + 2 ⇒    5x + 0 – z = 0      ....(4)
(3) × 3 ⇒ 3x + 3y + 3z = 81   ....(3)
(2) × 1 ⇒   x + 3y – 2z = 0     ....(2)
                 (–)   (–)   (+)                  
(3) – 2 ⇒     2x + 5z = 81      ....(5)
(4) × (5) ⇒ 25x – 5z = 0        ....(4)
(5) × 1 ⇒    2x + 5z = 81        ....(5)  
(4) + 5 ⇒         27x = 81
                           x = `81/27` = 3

Substituting the value of x in (5)

6 + 5z = 81

5z = 81 – 6

5z = 75

z = `75/5` = 15

Substituting the value of x = 3

and z = 15 in (3)

3 + y + 15 = 27

y + 18 = 27

y = 27 – 18

= 9

The value of x = 3, y = 9 and z = 15

This system of equations have unique solution.

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Simultaneous Linear Equations in Three Variables
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Chapter 3: Algebra - Exercise 3.1 [Page 93]

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Samacheer Kalvi Mathematics [English] Class 10 SSLC TN Board
Chapter 3 Algebra
Exercise 3.1 | Q 2. (iii) | Page 93
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