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Solve the Following Systems of Equations: `1/(5x) + 1/(6y) = 12` `1/(3x) - 3/(7y) = 8, X ~= 0, Y != 0` - Mathematics

Solve the following systems of equations:

`1/(5x) + 1/(6y) = 12`

`1/(3x) - 3/(7y) = 8, x ~= 0, y != 0`

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Solution

Taking 1/x = u and 1/y = v the given equations become

`u/5 + v/6 = 12`

`=> (6u + 5v)/30 = 12`

`=> 6u + 5v = 360`   .....(i)

And `u/3 - (3v)/7 = 8`

`=> (7u + 9v)/21 = 8`

=> 7u - 9v = 168 ....(ii)

Let us eliminate ‘v’ from equation (i) and (ii), Multiplying equation (i) by 9 and equation (ii) by 5, we get

54u + 45y = 3240 .......(iii)

35u - 45v = 840 .....(iv)

Adding equation (i) adding equation (ii), we get

54u + 35u = 3240 + 840

=> 89u = 4080

`=> u = 4080/89`

Putting u = 4080/89 in equation (i) we get

`6 xx 4080/89 + 5v = 360`

`=> 24480/89 + 5v = 360`

`=> 5v = 360 - (24480)/89`

`=> 5v = (32040 - 24480)/89`

`=> 5v = 7560/89`

`=> v = 7560/(5 xx 89)`

`=> c = 1512/89`

Hence  `x = 1/u = 89/4080` and `y = 1/v = 89/1512`

So, the solution of the given system of equation `x = 89/4080, y = 89/1512`

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

RD Sharma Class 10 Maths
Chapter 3 Pair of Linear Equations in Two Variables
Exercise 3.3 | Q 21 | Page 45
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