# What is the point of intersection of two lines 4x + 3y +7 =0 and 6x + 9y =0?

*print*Print*list*Cite

At the point of intersection the x and y values are equal.

Therefore we use the equations we have to find the point of intersection.

4x + 3y +7 =0… (1)

6x + 9y =0… (2)

Write (2) as 6x = -9y

=> x = - (3/2) y

Substitute x = - (3/2) y in (1)

4x + 3y +7 =0

=> 4*(-3/2) y + 3y +7 =0

=> -6y + 3y = -7

=> y = 7/3

Also, x = - (3/2) y = - (3/2)*(7/3) = -7/2

**Therefore we get the point of intersection as (-7/2, 7/3)**

Solve for x and y to find the point of intersection. (x,y)

Solve 6x+9y=0 for x in terms of y

6x=-9y

x=(-9y/6)

Substitute (-9y/6) for each x in the equation

4x+3y+7=0

4(-9y/6)+3y+7=0

Solve for y

Expand

-36y/6+3y+7=0

Simplify

-6y+3y+7=0

Combine like terms

-3y+7=0

Subtract 7 from both sides

-3y=-7

Divide both sides by -3

y=7/3

Substitute 7/3 for y in the equation 6x+9y=0

6x+9(7/3)=0

Solve for x

Expand

6x+63/3=0

Simplify

6x+21=0

Subtract 21 from both sides

6x=-21

Divide both sides by 3

2x=-7

Divide both sides by 2

x=-7/2

Point of intersection is (-7/2,7/3)

To find the point of intersection of the lines 4x+3y+7= 0 and 6x+9y = 0.

Solution:

4x+3y +7 = 0, we rewrite as below:

4x+3y = -7.........(1).

6x+3y = 0 which we simplify by dividing by 3:

2x+3y = 0.....(2).

So eq(1) - 2*eq(2) gives:

4x+3y - 2(2x+3y) = -7-2*0 = -7

3y-6y = -7.

-3y = -7.

y = -7/-3

y = 7/3

We put y = 7/3 in equation 2x+3y = 0:

2x+7 = 0.

2x= -7.

x= -7/2.

x = -7/2.

Therefore x = -7/2 and y = 7/3