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Solve the Following Systems of Equations: `4/X + 3y = 8` `6/X - 4y = -5` - Mathematics

Solve the following systems of equations:

4u + 3y = 8

`6u - 4y = -5`

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Solution

Taking `1/x = u` then given equations become

4u + 3y = 8 ......(i)

6u - 4y = -5 ..........(ii)

From (i), we get

4u = 8 -3y

`=> u = (8 - 3y)/4`

Substituting u = `(8 - 3y)/4` in (ii) we get

From (ii), we get

`6((8-3y)/4) - 4y = -5`

`=> (3(8 - 3y))/4 - 4y = -5`

`=> (24 - 9y)/2 - 4y= -5`

`=> (24 - 9y - 8y)/2 = -5`

=> 24 - 17y = -10

=> -17y = -10 -24

=> -17y = -34

`=> y = (-34)/(-17) = 2`

Putting y = 2 in u =`(8-3y)/4` we get

`u = (8 - 3xx 2)/4 = (8-6)/4 = 2/4 = 1/2`

Hence   x = 1/u = 2

So, the solution of the given system of equation is x  2, y

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APPEARS IN

RD Sharma Class 10 Maths
Chapter 3 Pair of Linear Equations in Two Variables
Exercise 3.3 | Q 8 | Page 44
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