[tex]\huge\mathfrak \red{solve \: the \: equations \: }[/tex]
✰ 4x – 18 = 3y , 6x + 7y – 4 = 0
✰ 5/x + 6y = 13 , 3

1 thought on “<br />[tex]\huge\mathfrak \red{solve \: the \: equations \: }[/tex]<br />✰ 4x – 18 = 3y , 6x + 7y – 4 = 0<br />✰ 5/x + 6y = 13 , 3”

1. Step-by-step explanation:

   Given:–

   Pair of linear equations are:

   1) 4x – 18 = 3y and 6x + 7y – 4 = 0

   2)5/x + 6y = 13 , 3/x + 4y = 7

   To find:–

   Solve the given equations ?

   Solution:–

   1)

   Given pair of linear equations are

   4x – 18 = 3y

   =>4x -18 -3y = 0

   =>4x -3y = 18———(1)

   on multiplying with 3 then

   =>12x-9y = 54 ——–(2)

   and 6x+7y -4 = 0

   =>6x +7y = 4

   on multiplying with 2 then

   =>12x +14y =8——–(3)

   On subtracting (3) from (2) then

   12x +14y = 8

   12x-9y = 54

   (-)

   __________

   0+23y = -46

   __________

   =>23y = -46

   =>y = -46/23

   =>y = -2

   In Substituting the value of y in (1) then

   4x -3(-2) = 18

   =>4x +6 = 18

   =>4x = 18-6

   =>4x = 12

   =>x = 12/4

   =>x = 3

   The value of x = 3 and y=-2

   Solution = (3,-2)

   Verification:–

   If x = 3 and y=-2 then

   4(3)-3(-2)=13+6=18

   6(3)+7(-2)=18-14=4

   Verified the given relations

   2)

   Given equation are: 5/x + 6y = 13———(1)

   3/x + 4y = 7——-(2)

   Put 1/x = a then

   5a+6y = 13——-(3)

   on multiplying with 3 then

   15a+18y = 39————(4)

   and 3a+4y = 7——-(5)

   on multiplying with 5 then

   15a+20y = 35———–(6)

   On Subtracting (6) from (5) then

   15a +20 y = 35

   15a +18y = 39

   (-)

   ____________

   0+2y = -4

   ____________

   =>2y = -4

   =>y = -4/2

   y = -2

   On Substituting the value of y in (3)

   =>5a+6(-2)=13

   =>5a-12 = 13

   =>5a = 13+12

   =>5a = 25

   =>a = 25/5

   =>a = 5

   =>a=1/x = 5

   =>x = 1/5

   The value of x = 1/5 and y = -2

   The solution=(1/5 ,-2)

   Verification:–

   If x = 1/5 and y = -2 then

   5/(1/5)+6(-2)

   =>25-12= 13

   and

   3/(1/5)+4(-2)

   =>15-8 = 7

   Verified the given relations.

   Solutions:–

   Solutions are I) (3,-2) and ii)(1/5,-2)

   Used method :–

   Elimination method.

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