Tamil Nadu Board of Secondary EducationSSLC (English Medium) Class 10th
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Solve the following system of linear equations in three variables 1x-2y+4=0;1y-1z+1=0;2z+3x=14 - Mathematics

Sum

Solve the following system of linear equations in three variables

`1/x - 2/y + 4 = 0; 1/y - 1/z + 1 = 0; 2/z + 3/x = 14`

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Solution

Let `1/x` = p, `1/y` = q and `1/z` = r

p – 2q + 4 = 0

p – 2q = – 4   ...(1)

q – r + 1 = 0

q – r = – 1  ...(2)

3p + 2r = 14  ...(3)

(1) × 1 ⇒    p – 2q + 0 =  – 4  ...(1)
(2) × 2 ⇒    0 + 2q – 2r = – 2 ...(2)
(1) + (2) ⇒  p + 0  – 2r = – 6
                          p – 2r = – 6   ...(4)
        (3) ⇒       3p + 2r = 14    ...(3)
(3) + (4) ⇒       4p + 0 = 8
                              4p = 8
                                p = `8/4` = 2 

Substituting the value of p = 2 in (1)

2 – 2q = – 4

– 2q = – 4 – 2

– 2q = – 6

q = `6/2` = 3

Substituting the value of q = 3 in (2)

3 – r = 1

– r = – 1 – 3

r = 4

But `1/x` = p

`1/x` = 2

2x = 1

 x = `1/2`

`1/y` = q

`1/y` = 3

3y = 1

y = `1/3`

`1/z` = r

`1/z` = 4

4y = 1

z = `1/4`

The value of x = `1/2`, y = `1/3` and z = `1/4`

  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. (ii) | Page 92
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