नानक,O Sv-4ye8=0T By 9-0

1.	नानक,O Sv-4ye8=0T By 9-0
Answer» 5x-4y+8=0 be equation 1st 7x+6y-9=0 be equation 2nd solve these two equations to find the intersection point of these two equation 5x-4y=-8be equation 1st 7x+6y=9 be equation 2nd multiply 1st equation by 3 and multilply 2ndequation by 2 to get the coefficient of y equal using elimination method to solve x and y 5x(3)-4y(3) = -8×37x(2)+6y(2) = 9×2 15x-12y=-24 equation 3rd 14x+12y=18 equation 4th add these two equation 15x+14x-12y+12y=-24+18 29x=-6 x=-6/29 substitute x=-6/29 equation 1st 5x-4y=-8 5(-6/29)+-4y=-8 4y=(30/29)-8 4y=(30-29×8)/29 4y=-202/29 2y=-101/29 y=-101/58 we get the value of x and y now on representing these two line on graph i putthese two lines intersect at point p point p is (-6/29,-101/58) Like my answer if you find it useful!

Answer»

5x-4y+8=0 be equation 1st

7x+6y-9=0 be equation 2nd

solve these two equations to find the intersection point of these two equation

5x-4y=-8be equation 1st

7x+6y=9 be equation 2nd

multiply 1st equation by 3 and multilply 2ndequation by 2 to get the coefficient of y equal

using elimination method to solve x and y

5x(3)-4y(3) = -8×37x(2)+6y(2) = 9×2

15x-12y=-24 equation 3rd

14x+12y=18 equation 4th

add these two equation

15x+14x-12y+12y=-24+18

29x=-6

x=-6/29

substitute x=-6/29 equation

1st 5x-4y=-8

5(-6/29)+-4y=-8

4y=(30/29)-8

4y=(30-29×8)/29

4y=-202/29

2y=-101/29

y=-101/58

we get the value of x and y now on representing these two line on graph i putthese two lines intersect at point p point p is (-6/29,-101/58)

Like my answer if you find it useful!

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