# In which quadrant does the line 5y-3x+1 = 0 meets the line 2y-4x +7 = 0?

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To find the point of intersection we need to solve the system of equations

5y - 3x + 1 = 0 ...(1)

2y - 4x + 7 = 0 ...(2)

2*(1) - 5*(2)

=> 10y - 6x + 2 - 10y + 20x - 35 = 0

=> 14x - 33 = 0

=> x = 33/14

y = 17/14

As both x and y are positive the lines intersect in the first quadrant.

**The line 5y-3x+1 = 0 meets the line 2y-4x +7 = 0 in the first quadrant.**

To determine the quadrant we will need to find the intersection point between the lines , then we determine the location.

==> 5y-3x + 1 = 0 ==> y= (3x-1)/5 ............(1)

==> 2y-4x + 7 = 0==> y= (4x-7)/2 ............(2)

Now we will find the intersection point.

==> (3x-1)/5 = (4x-7)/2

Cross multiply.

==> 2(3x-1) = 5(4x-7)

==> 6x -2 = 20x - 35

==> 14x = 33

==> x = 33/14= 2.36

==> y= (3x-1)/5 = (3*2.36 -1 )/5 = 1.21.

Then the point of intersection is ( 2.36, 1.21)

**Then, the point if located in the first quadrant.**