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# Solve 2/(x+2y)+6/(2x−y)=4 ;5/2(x+2y)+1/(3(2x−y))=1 where, x + 2y ≠ 0 and 2x – y ≠ 0 - CBSE Class 10 - Mathematics

ConceptEquations Reducible to a Pair of Linear Equations in Two Variables

#### Question

Solve \frac{2}{x+2y}+\frac{6}{2x-y}=4\text{ ;}\frac{5}{2( x+2y)}+\frac{1}{3( 2x-y)}=1 where, x + 2y ≠ 0 and 2x – y ≠ 0

#### Solution

Taking \frac { 1 }{ x+2y } = u and \frac { 1 }{ 2x-y } = v , the above system of equations becomes

2u + 6v = 4 ….(1)

\frac { 5u }{ 2 } + \frac { v }{ 3 } = 1 ….(2)

Multiplying equation (2) by 18, we have;

45u + 6v = 18 ….(3)

Now, subtracting equation (3) from equation (1), we get ;

–43u = – 14 ⇒ u = \frac { 14 }{ 43 }

Putting u = 14/43 in equation (1), we get

2 × \frac { 14 }{ 43 } + 6v = 4

⇒ 6v = 4 – \frac { 28 }{ 43 } = \frac { 172-28 }{ 43 } ⇒ v = \frac {144 }{ 43 }

Now,

u = \frac { 14 }{ 43 } = \frac { 1 }{ x+2y }

⇒ 14x + 28y = 43 ….(4)

And,

v = \frac { 144 }{ 43 } = \frac { 1 }{ 2x-y }

⇒ 288x – 144y = 43 ….(5)

Multiplying equation (4) by 288 and (5) by 14, the system of equations becomes

288 × 14x + 28y × 288 = 43 × 288

288x × 14 – 144y × 14 = 43 × 4

⇒ 4022x + 8064y = 12384 ….(6)

4022x – 2016y = 602 ….(7)

Subtracting equation (7) from (6), we get

10080y = 11782 ⇒ y = 1.6(approx)

Now, putting 1.6 in (4), we get,

14x + 28 × 1.6 = 63

⇒ 14x + 44.8 = 63 ⇒ 14x = 18.2

⇒ x = 1.3 (approx)

Thus, solution of the given system of equation is x = 1.3 (approx), y = 1.6 (approx).

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Solution Solve 2/(x+2y)+6/(2x−y)=4 ;5/2(x+2y)+1/(3(2x−y))=1 where, x + 2y ≠ 0 and 2x – y ≠ 0 Concept: Equations Reducible to a Pair of Linear Equations in Two Variables.
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