Let d1 and d2 be two lines such that:

d1 : 4x + 3y =0

d2: 7x + 5y + 1 = 0

Both lines are intersecting at one point , then the point of intersection sould verify both equations. Therefore te point is a common solution for the system.

Then we need to solve the system:

Let us use the elemination method:

Multiply (1) by -5 and multiply (2) by 3:

-20x - 15 y = 0.........(1)

21x + 15x = -3 ..........(2)

**==> x = -3**

Now to find y, we will substitue with (1):

4x + 3y = 0

==> y= (-4/3)*x = -4/3 * -3 = 4

**==> y= 4**

**Then both lines intersects at the point (-3, 4)**

To find the point of intersection of

4x+3y=0..............(1)

7x+5y+1 = 0.........(2)

(1)*5 -(2)*3 eliminates y and gives:

4x*5-7x*3 -1*3 = 0

20x-21x = 3

-x = 3.

x = -3

Put x=3 in (1):

4(-3) +3y = 0

-12+3y = 0

3y = 12

y = 4.

So x= -3 and y = 4

The point of intersection of 4x+ 3y =0 and 7x + 5y +1 =0 is the point (x,y) which if substituted into either of the equations provides a solution

Now let’s take the equations:

4x+ 3y =0 …(1)

7x + 5y +1 =0 …(2)

From (1)

4x+ 3y =0

=> x = -3y/4

substituting x = -3y/4 in (2)

7*(-3y/4) + 5y+1 =0

=> -21y /4 + 20y/4 + 4/4 =0

=> -21 +20y +4 =0

=> -y = -4

=> y=4

Using y =4 in x = -3y/4

=> x = -3*4/4 = -3

**Therefore the point of intersection is (-3, 4)**