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Solve the following systems of equations: `1/(2(x + 2y)) + 5/(3(3x - 2y)) = (-3)/2` `5/(4(x + 2y)) - 3'/(5(3x - 2y)) = 61/60` - Mathematics

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प्रश्न

Solve the following systems of equations:

`1/(2(x + 2y)) + 5/(3(3x - 2y)) = (-3)/2`

`5/(4(x + 2y)) - 3'/(5(3x - 2y)) = 61/60`

उत्तर

Let `1/(x + 2y) = u and 1/(3x - 2y) = v`

Then, the given system of equation becomes

`u/2 + (5v)/3 = (-3)/2`

`=> (3u + 10v)/6 = (-3)/2`

`=> 3u + 10v =  (-3 xx 6)/2`

=> 3u + 10v = -9 ....(i)

And `(5u)/4 - (3v)/5 = 61/60`

`=> (25u - 12v)/20 = 61/60`

`=> 25u- 12v = 61/3`   ...(ii)

Multiplying equation (i) by 12, and equation (ii) by 10, we get

36u + 120vg = -108 ...(iii)

`250u - 120v = 610/3` .....(iv)

Adding equation (iii) and equation (iv), we get

`36u + 250uy = 610/3 - 108`

`=> 286u = (610 - 324)/3`

`=> 286u = 286/3`

=> u = 1/3

Putting u = 1/3 in equation (i) we get

`3 xx 1/3 + 10v = -9`

`=> 1 + 10v = -9`

=> 10v = -9 - 1

`=> v = (-10)/10 = -1`

Now `u = 1/(x + 2y)`

`=> 1/(x + y) = 1/3`

=> x + 2y = 3 ....(v)

And `v = 1/(3x - 2y)`

=> 1/(3x - 2y) = -1

=> 3x - 2y = -1 ......(vi)

Putting x = 1/2 in equation (v) we get

`1/2 + 2y = 3`

`=> 2y = 3 - 1/2`

`=> 2y = (6 -1)/2`

`=> y = 5/4`

Hence, solution of the given system of equations is x = 1/2, y = 5/4

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Pair of Linear Equations in Two Variables - Exercise 3.3 [पृष्ठ ४५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 10
अध्याय 3 Pair of Linear Equations in Two Variables
Exercise 3.3 | Q 32 | पृष्ठ ४५

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