# What is the point of intersection of the lines x + 6y = 7, 8x + y = 1 and 4x + 5y = 0?

### 3 Answers | Add Yours

We have the equations of three lines: x + 6y = 7, 8x + y = 1 and 4x + 5y = 0.

Now, we need to find their points of intersection.

- x + 6y = 7 and 8x + y = 1

x + 6y = 7

=> x = 7 – 6y

substituting this in 8x + y = 1

8*(7 – 6y) + y = 1

=> 56 – 48y + y = 1

=> 56 – 47y = 1

=> 47y = 55

=> y = 55/47

x = 7 – 6(57/47) = -1/47

We have the point (-1/47, 55/47)

- 8x + y = 1 and 4x + 5y = 0

8x + y = 1

=> y = 1 – 8x

Substituting this in 4x + 5y = 0

4x + 5(1 – 8x) =0

=> 4x + 5 – 40x = 0

=> 36x = 5

=> x = 5/36

y = 1- 8(5/36) = -1/9

The second point is (5/36, -1/9)

- 4x + 5y = 0 and x + 6y = 7

x + 6y = 7

=> x = 7 – 6y

Substituting this in 4x + 5y =0

=> 4(7 – 6y) + 5y = 0

=> 28 – 24y + 5y =0

=> 28 = 19y

=> y = 28/ 19

x = 7 – 6(28/19) = -35/19

The third point is (-35/19, 28/ 19)

**Therefore the points of intersection are (-1/47, 55/47), (5/36, -1/9) and (-35/19, 28/19)**

Given the lines

x + 6y = 7............(1)

8x + y = 1............(2)

4x + 5y = 0............(3)

We need to find the intersection point of all lines.

Then we need to determine the point the verifies all 3 equations.

First we will find the intersection point of the first two lines, then check if it verifies the third line.

We will write (1):

x = 7 - 6y

Then we will substitute into (2).

=> 8x + y = 1

==> 8(7-6y) + y = 1

==> 56 - 48y + y = 1

==> - 47y = - 55

==> y= 55/47

==> x = 7 - 6y = 7 - 6(55/47) = 7- 330/ 47 = - 1/47

==> x= -1/47

Now we will check if the point ( -1/47, 55/47) verifies equation(3).

==> 4x + 5y = 0

==> 4( -1/47) + 5(55/47) = 0

==> -4/47 + 275/47 = 0

==> 251/47 = 0

**Then, the three lines do not intersect at one point.**

I needed the point of intersection of all the three lines with each other.