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

Solve 13 (x + y – 5) = y – z = 2x – 11 = 9 – (x + 2z) - Mathematics

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Question

Solve `1/3` (x + y – 5) = y – z = 2x – 11 = 9 – (x + 2z)

Sum
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Solution

`1/3` (x + y – 5) = y – z 

x + y – 5 = 3y – 3z

x + y – 3y + 3z = 5

x – 2y + 3z = 5  ...(1)

y – z – 2x – 11

– 2x + y – z = – 11

2x – y + z = 11  ...(2)

2x – 11 = 9 – (x + 2z)

2x – 11 = 9 – x – 2z

2x + x + 2z = 9 + 11

3x + 2z = 20  ...(3)

(1) × 1  ⇒   x – 2y + 3z = 5     ....(1)
(2) × 2  ⇒ 4x – 2y + 2z = 22   ....(4)
                (–)   (+)    (–)    (–)        
(1) – (4) ⇒ – 3x + 0 + z = – 17

3x – z = 17  ....(5)
Subtracting (3) and (5) we get
(3) ⇒ 3x + 2z = 20
(5) ⇒ 3x –   z = + 17
        (–)    (+)     (–)  
         3z = 3
           z = `3/3` = 1 

Substitute the value of z = 1 in (3)

3x + 2(1) = 20

3x = 20 – 2

3x = 18

x = `18/3` 

= 6

Substitute the value of x = 6, z = 1 in (2)

2(6) – y + 1 = 11

12 – y + 1 = 11

13 – y = 11

– y = 11 – 13

– y = – 2

y = 2

∴ The value of x = 6, y = 2 and z = 1

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Chapter 3: Algebra - Unit Exercise – 3 [Page 156]

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Samacheer Kalvi Mathematics [English] Class 10 SSLC TN Board
Chapter 3 Algebra
Unit Exercise – 3 | Q 1 | Page 156

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