Simultaneous Equation...!

I need help on this as i am having difficulties to solve it. Can someone teach me how to do this?

2y=3x-1
4x/y + 9y/x=15

2y = 3x − 1
y = (3x−1)/2

Replace y in second equation with (3x−1)/2, then solve for x

4x/((3x−1)/2) + 9((3x−1)/2)/x = 15
8x/(3x−1) + 9(3x−1)/(2x) = 15

Multiply both sides by 2x(3x−1)

8x(2x) + 9(3x−1)(3x−1) = 15(2x)(3x−1)
16x² + 9(9x² − 6x + 1) = 15 (6x² - 2x)
16x² + 81x² − 54x + 9 = 90x² − 30x
16x² + 81x² − 90x² − 54x + 30x + 9 = 0
7x² − 24x + 9 = 0
7x² − 21x − 3x + 9 = 0
7x (x − 3) − 3 (x − 3) = 0
(x − 3) (7x − 3) = 0

x = 3 -----> y = (3x−1)/2 = (9−1)/2 = 4
x = 3/7 -----> y = (3x−1)/2 = (9/7−1)/2 = 1/7

Solutions: {(3, 4), (3/7, 1/7)}

You need to rearrange the first equation so that y= (3x-1)/2
Then you can put the answer for y into the second equation, so the second equation looks like this:
(4x/3x-1)/2 + 9(3x-1)/2 = 15
You can then cancel it down so that it gives you the x value which by working it out quickly (don't trust my answer) is x=3
Once you know your x value (i'm using 3 for now) you put that back into the first equation which was y= (3x-1)/2.
So now you have y= ((3x3)-1)/2 which is (9-1)/2 which is 4,
SO: if my y value was correct (you have to do it yourself:) ), the x value is 3 and the y value is 4.

Good luck :)

2y=3x-1
4x/y + 9y/x=15
first of all , both x and y must not be 0
then we can proceed:
first equation:
2y = 3x - 1
y = 3/2 x - 1/2

second
4x/y + 9y/x=15
mcm is x*y...
4x² + 9y² = 15xy

let's combine both:
4x² + 9(3/2 x - 1/2)² = 15(3/2 x - 1/2)x
4x² + 9(9/4 x² - 3/2x + 1/4) = 45/2 x² - 15/2 x
4x² + 81/4 x² - 27/2 x + 9/4 = 45/2 x² - 15/2 x
all * 4
16x² + 81x² - 54x + 9 = 90x² - 30x
7x² - 24x + 9 = 0
factor it..
(x - 3)(7x - 3) = 0
x ₁ ₂= +3 ; +3/7
y₁ = 3/2 * +3 - 1/2 = 4
y₂ = 3/2 * 3/7 - 1/2 = 1/7
solution is:
A(3,4) , B(3/7 , 1/7)

2y=3x-1
y = (3x-1)/2 ......(1)

4x/y + 9y/x=15
4x^2 + 9y^2 = 15xy ..........(2)

Plug equation (1) into (2);
4x^2 + 9 [(3x-1)/2 ]^2 = 15x [(3x-1)/2 ]
:
:
:
7x^2 - 24x + 9 = 0
(7x-3)(x-3) = 0

x = 3/7, x = 3

x = 3/7, y = [3(3/7)-1]/2 = 1/7

x = 3, y = [3(3)-1]/2 = 4