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Solve the following system of equations by using the method of elimination by equating the co-efficients.

`\frac { x }{ y } + \frac { 2y }{ 5 } + 2 = 10; \frac { 2x }{ 7 } – \frac { 5 }{ 2 } + 1 = 9`

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#### Solution

The given system of equation is

`\frac { x }{ y } + \frac { 2y }{ 5 } + 2 = 10 ⇒ \frac { x }{ y } + \frac { 2y }{ 5 } = 8 …(1)`

`\frac { 2x }{ 7 } – \frac { 5 }{ 2 } + 1 = 9 ⇒ \frac { 2x }{ 7 } –\frac { 5 }{ 2 } = 8 ….(2)`

The equation (1) can be expressed as :

`\frac { 5x+4y }{ 10 } = 8 ⇒ 5x + 4y = 80 ….(3)`

Similarly, the equation (2) can be expressed as :

`\frac { 4x-7y }{ 14 } = 8 ⇒ 4x – 7y = 112 ….(4)`

Now the new system of equations is

5x + 4y = 80 ….(5)

4x – 7y = 112 ….(6)

Now multiplying equation (5) by 4 and equation (6) by 5, we get

20x – 16y = 320 ….(7)

20x + 35y = 560 ….(8)

Subtracting equation (7) from (8), we get ;

`y = -\frac { 240 }{ 51 }`

Putting `y = -\frac { 240 }{ 51 } ` in equation (5), we get ;

`5x + 4 × \frac { -240 }{ 51 } = 80 ⇒ 5x – \frac { 960 }{ 51 } = 80`

`⇒ 5x = 80 + \frac { 960 }{ 51 } = \frac { 4080+960 }{ 51 } = \frac {5040 }{ 51 }`

`⇒ x = \frac { 5040 }{ 255 } = \frac { 1008 }{ 51 }= \frac { 336 }{ 17 }`

`⇒ x = \frac { 336 }{ 17 }`

Hence, the solution of the system of equations is, `x = \frac { 336 }{ 17}, y = -\frac { 80 }{ 17 }`

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**Read the following passage:**

Two schools 'P' and 'Q' decided to award prizes to their students for two games of Hockey ₹ x per student and Cricket ₹ y per student. School 'P' decided to award a total of ₹ 9,500 for the two games to 5 and 4 Students respectively; while school 'Q' decided to award ₹ 7,370 for the two games to 4 and 3 students respectively. |

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