मराठी

Solve the following system of equations: sqrt(2)x + sqrt(3)y = 5 sqrt(3)x – sqrt(8)y = –sqrt(6)

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प्रश्न

Solve the following system of equations:

`sqrt(2)x + sqrt(3)y = 5`

`sqrt(3)x - sqrt(8)y = -sqrt(6)`

बेरीज
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उत्तर

Given: `sqrt(2)x + sqrt(3)y = 5, sqrt(3)x - sqrt(8)y = -sqrt(6)`

Step-wise calculation:

1. Simplify `sqrt(8) = 2sqrt(2)` 

So the system is `sqrt(2)x + sqrt(3)y = 5`   ...(1)

`sqrt(3)x - 2sqrt(2)y = -sqrt(6)`   ...(2)

2. Eliminate x: multiply (1) by `sqrt(3)` and (2) by `sqrt(2)`: 

`(1)·sqrt(3): sqrt(6)x + 3 y = 5sqrt(3)`

`(2)·sqrt(2): sqrt(6)x - 4 y = -2sqrt(3)`

3. Subtract the second from the first:

`(sqrt(6) x + 3y) − (sqrt(6) x − 4y) = 5 sqrt(3) − (−2 sqrt(3))`

⇒ `7y = 7sqrt(3)` 

⇒ `y = sqrt(3)`

4. Substitute `y = sqrt(3)` into (1): 

`sqrt(2)x + sqrt(3) xx sqrt(3) = 5` 

⇒ `sqrt(2)x + 3 = 5` 

⇒ `sqrt(2)x = 2` 

⇒ `x = 2/sqrt(2)`

⇒ `x = sqrt(2)`

5. Check In (2): `sqrt(3) xx sqrt(2) - 2 sqrt(2) xx sqrt(3)` 

= `sqrt(6) - 2sqrt(6)`

= `-sqrt(6)`, matches RHS.

`x = sqrt(2), y = sqrt(3)`

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पाठ 3: Pair of Linear Equations in Two Variables - EXERCISE 3.3 [पृष्ठ ३.२९]

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आर.डी. शर्मा Mathematics [English] Class 10
पाठ 3 Pair of Linear Equations in Two Variables
EXERCISE 3.3 | Q 25. | पृष्ठ ३.२९
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