10 points! MATH LOVERS! ACT type questions please show your work

find the equation of a line that passes thru (9,-6) and is parallel to y=5/3x-14

find the equation of a line passes thru (-1,4) and is perpindicular to y=1/5x+2

what is the vertex of the graph of y=lx-9l-4? (buy the way that IS NOT the number 91 <----- its like parathesis thing)

what is the value of x in the system of equations
{3x-y=13
{5x+2y=18


which ordered pair is a solution to the system
{y=2x-11
{x+4y=-17

Given line's slope is 5/3. A parallel line would have the same slope. Use point-slope formula:
y - (-6) = 5/3(x - 9)
y + 6 = 5/3x - 15
y = 5/3 x - 21

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Same process here, except that a perpendicular line's slope would be the opposite reciprocal of 1/5, which would be -5.
y - 4 = -5(x - (-1))
y = -5x - 1

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y = |x - 9| - 4
Vertex would occur at (9, -4)

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3x - y = 13
5x + 2y = 18

Multiply the first equation by 2 to eliminate y:
6x - 2y = 26
5x + 2y = 18
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11x = 44
x = 4

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y = 2x - 11
x + 4y = -17

By substitution:
x + 4(2x - 11) = -17
x + 8x - 44 = -17
9x = 27
x = 3

y = 2x - 11
y = 2(3) - 11
y = 6 - 11
y = -5

(x, y) = (3, -5)

slope=(5/3) then the equation is
(y+6)=(5/3)(x-9)
y+6=(5/3)x -15
y=(5/3)x-21

the slope=-5 then the equation is

(y-4)=-5(x+1)
y-4=-5x-5
y=-5x-1

the vertex is (9,-4)

3x-y=13 eq 1
5x+2y=18 eq 2

11x=44 eq 3= eq 2+2eq 1
x=44/11
x=4
y=3x-13 from eq 1
y=12-13
y=-1

y=2x-11
x+4y=-17
x+4(2x-11)=-17
x+8x-44=-17
9x=44-17
9x=27
x=27/9
x=3
y=2x-11
y=6-11
y=-5

y=5/3x -21
y=-5x -1
(9,-4)
x=4
(3,-5)