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