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# Solve the following systems of equations: 6/(x + y) = 7/(x - y) + 3 1/(2(x + y)) = 1/(3(x - y)), where x + y ≠ 0 and x – y ≠ 0 - CBSE Class 10 - Mathematics

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ConceptAlgebraic Methods of Solving a Pair of Linear Equations Substitution Method

#### Question

Solve the following systems of equations:

6/(x + y) = 7/(x - y) + 3

1/(2(x + y)) = 1/(3(x - y)), where x + y ≠ 0 and x – y ≠ 0

#### Solution

Let 1/(x + y) = u and 1/(x - y) = v Then, the given system of equation becomes

6u = 7v + 3

=> 6u - 7v = 3 .....(i)

And u/2 = v/3

=> 3u = 2v

=> 3u - 2v = 0 ......(ii)

Multiplying equation   ii by 2, and equation (i) by 1, we get

6u - 7v = 3 ....(iii)

6u - 4v = 0 ......(iv)

Subtracting equation (iv) from equation (iii), we get

-7 + 4v = 3

=> -3v = 3

=> v = -1

Puttting v = -1 in equation (ii), we get

3u - 2xx (-1) = 0

=> 3u + 2 = 0

=> 3u = -2

=> u = (-2)/3

Now u = (-2)/3

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

=> x + y = (-3)/2   ...(v)

And v= -1

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

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

Adding equation (v) and equation (vi), we get

2x = (-3)/2 - 1

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

=> 2x = (-5)/2

=> x = (-5)/4

Putting x = (-5)/4  in equation (vi), we get

(-5)/4 - y = -1

=> (-5)/4 + 1 = y

=> (-5+4)/4 = y

=> (-1)/4 = y

=> y = (-1)/4

Hence, solution of the system of equation is x = (-5)/4, y = (-1)/4

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Solution Solve the following systems of equations: 6/(x + y) = 7/(x - y) + 3 1/(2(x + y)) = 1/(3(x - y)), where x + y ≠ 0 and x – y ≠ 0 Concept: Algebraic Methods of Solving a Pair of Linear Equations - Substitution Method.
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