Tamil Nadu Board of Secondary EducationSSLC (English Medium) Class 10th
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Solve the following system of linear equations in three variables x + 20 = 3y2+10 = 2z + 5 = 110 – (y + z) - Mathematics

Sum

Solve the following system of linear equations in three variables

x + 20 = `(3y)/2 + 10` = 2z + 5 = 110 – (y + z)

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Solution

x + 20 = `(3y)/2 + 10` 

Multiply by 2

2x + 40 = 3y + 20

2x – 3y = – 40 + 20

2x – 3y = – 20   ...(1)

`(3y)/2 + 10` = 2z + 5

Multiply by 2

3y + 20 = 4z + 10

3y – 4z = 10 – 20

3y – 4z = – 10   ...(2)

2z + 5 = 110 – (y + z)

2z + 5 = 110 – y – z

y + 3z = 110 – 5

y + 3z = 105   ...(3)

(3) × (3) ⇒   3y + 9z = 315   ...(3)
(2) × (1) ⇒   3y – 4z = – 10   ...(2)
                    (–) (+)     (+)            
(3) – (2) ⇒          13z = 325
                              z = `325/13` = 25

Substitute the value of z = 25 in (2)

3y – 4(25) = – 10

3y – 100 = – 10

3y = – 10 + 100

3y = 90

y = `90/3` = 30

∴ The value of x = 35, y = 30 and z = 25

Substitute the value of y = 30 in (1)

2x – 3(30) = – 20

2x – 90 = – 20

2x = – 20 + 90

2x = 70

x = `70/2` = 35

  Is there an error in this question or solution?
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APPEARS IN

Samacheer Kalvi Mathematics Class 10 SSLC Tamil Nadu State Board
Chapter 3 Algebra
Exercise 3.1 | Q 1. (iii) | Page 92
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