Question

3x-5y-4=0 and 9x=2y+7 solve eqn. By elimination method

Answer 1

Answer:

The solution of the given simultaneous equations is ( x, y ) = [ ( 9 / 13 ), ( - 5 / 13 ) ].

Step-by-step-explanation:

The given simultaneous equations are

3x - 5y - 4 = 0 and 9x = 2y + 7.

Now,

3x - 5y - 4 = 0

⇒ 3x - 5y = 4 - - ( 1 )

Now,

9x = 2y + 7

⇒ 9x - 2y = 7 - - ( 2 )

Multiplying equation ( 1 ) by 3, we get,

3x - 5y = 4 - - ( 1 )

⇒ 3 * ( 3x - 5y ) = 4 * 3

⇒ 9x - 15y = 12 - - ( 3 )

By subtracting equation ( 2 ) from equation ( 3 ), we get,

9x - 15y - ( 9x - 2y ) = 12 - 7

⇒ 9x - 15y - 9x + 2y = 5

⇒ 9x - 9x - 15y + 2y = 5

⇒ 0 - 13y = 5

⇒ - 13y = 5

⇒ y = - 5 / 13

Now, by substituting y = - 5 / 13 in equation ( 1 ), we get,

3x - 5y = 4 - - ( 1 )

⇒ 3x - 5 * ( - 5 / 13 ) = 4

⇒ 3x + 25 / 13 = 4

⇒ 3x = 4 - ( 25 / 13 )

⇒ 3x = ( 4 * 13 - 25 ) / 13

⇒ 3x = ( 52 - 25 ) / 13

⇒ 3x = 27 / 13

⇒ x = 27 / 13 * 1 / 3

⇒ x = 27 ÷ 3 * 1 / 13

⇒ x = 9 / 13

∴ The solution of the given simultaneous equations is ( x, y ) = [ ( 9 / 13 ), ( - 5 / 13 ) ].

Answer 2

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(i) We have, 3x−5y−4=0

⇒3x−5y=4...(i)

Again 9x=2y+7

⇒9x−2y=7...(ii)

By Elimination Method:

Multiplying equation (i) by 3, we get

9x−15y=12...(iii)

Subtracting (ii) from (iii), we get

9x−15y=12

9x−2y = 7 / −13y=5

⇒y=− 5/13

Putting the value of equation (ii), we get

9x - 2 (-5/13) = 7

9x + 10/13 = 7

9x = 7 - 10/13

9x = 91 - 10 / 13

9x = 81/13

x = 9/13

Hence , the required solution is

x = 9/13

y = -5/13

By Substitution Method:

Expressing x in terms of y from equation (i), we have

x = 4 + 5y / 3

Substituting the value of x in equation (ii), we get

9 × ( 4 + 5y / 3 ) - 2y = 7

3 × ( 4 + 5y ) - 2y = 7

12 + 15y - 2y = 7

13y =7−12

y = -5/13

Putting the value of y in equation (i), we have

3x - 5 × ( -5/13 ) = 4

3x + 25/13 = 4

3x = 4 - 25/13

3x = 27/13

x = 9/13

Hence , the required solution is

x = 9/13

y = -5/13

(ii) We have, 2/x + 2y/2 =−1

3x + 4y / 6 = -1

∴3x+4y = −6 ...(i)

& x - y/2 = 3 => 3x - y / 3 = 3

∴3x−y =9 ...(ii)

By Elimination Method:

Subtracting (ii) from (i), we get

5y = -15

y = -15 / 5

y = -3

Putting the value of y in equation (i), we get

3x + r × (−3) = −6

⇒3x−12=−6

⇒3x = −6 + 12

⇒3x = 6

Hence, Solution is

x = 2

y = −3

By Substitution Method:

Expressing x in terms of y from equation (i), we have

x = -6 -4y / 3

Substituting the value of x in equation (ii) from equation (i), we get

3 × ( -6 -4y / 3 ) -y = 9

⇒−6 −4y −y =9

⇒−6 −5y =9

⇒−5y =9+6 => 15

∴ y = 15/-5 = -3

Putting the value of y in equation (i), we get

3x + × (−3) = −6

⇒3x − 12 = −6

⇒3x = 12 − 6 = 6

x = 6/3 = 2

Hence, the required solution is

x=2

y=−3