What did I do wrong??? (maths)?

Question

Ok here is the question: Find the equation of the line which is parallel to the line 3x-2y+5=0 and which passes through the point of intersection of the lines 4x+y+3=0 and 5x-2y+8=0

My working:
m=3/2 <--- from 3x-2y+5

(4x+y+3)+k(5x-2y+8)
4x+y+3+5kx-2ky+8k
x(4+5k)+y(1-2k)+(3+8k)
4+5k/2k-1=3/2 <---- gradient of line above
8+10k=6k-3
4k=-11
k=-11/4
(4x+y+3)-11/4(5x-2y+8)
4(4x+y+3)-11(5x-2y+8)
16x+4y+12-55x+22y-88
= -39x+26y-76=0 <------ my answer, however this is wrong and instead it is supposed to be 39x-26y+76=0 according to my textbook.

Where did i go wrong?

Answer 1

y = - 3 - 4x
5x - 2(- 3 - 4x) + 8 = 0
5x + 6 + 8x = - 8
13x = - 14
x = - 14/13

y = - 3 + 56/13
y = - 39/13 + 56/13
y = 17/13

Point of intersection is (-14/13 , 17/13)

3x - 2y = -5
2y = 3x + 5
y = (3/2) x + 5/2
m = 3/2

Required line:-
y - 17/13 = (3/2) (x + 14/13)
y = (3/2) x + 42/26 + 34/26
y = (3/2) x + 76/26
26y = 39x + 76
39x - 26y + 76 = 0
which IS THE SAME AS
-39x + 26y - 76 = 0
(multiply both sides by (-1) )
You have done the question!!!

Answer 2

Same thing. Multiply both sides by -1 and you have the textbook answer. You are right.

Answer 3

I've never seen it done this way. You were close, you must have dropped a negative somewhere. I would have just used a system of equations and found the point of intersection using substitution or elimination. This way you have the slope and a point on the line.

Answer 4

you were quite right.

in the textbook, the equation was multiplied throughout by -1.

both answers are right.