How to do these two math problems? (Show work) Please and thank you!?

Question

1. What is the equation of a line that passes through the point (-3,-11) and is parallel to the line whose equation is 2x-y=4?

2. Write the slope-intercept form of the equation of a line passsing through the point (-3,0) and parallel to the line y=6x+3

Answer 1

1. Given point ( -3,-11) and the equation is 2x - y = 4

The parallel line equation for the given equation 2x - y = 4 is 2x - y =k ----------(1)

Equation (1) passes through the point ( -3 , -11)

Say ( x,y) = ( -3,-11)

Substitute ( x,y) = (-3,-11) in (1)

2(-3) - (-11) = k

-6+11 = k

5 = k

Therefore k = 5

Substitute in (1)

2x - y = 5

Therefore parallel line equation is 2x - y = 5 

Answer 2

1. The line equation is 2x - y = 4

Rewrite the equation

y = 2x - 4

Compare the equation with y = mx + b here m is slope and b is y intercept.

If the two lines are parallel then slopes are same (m1 = m2 = 2)

The point is (-3, -11)

(Substitute x = -3 and y = -11)

-11 = 2(-3) + b

b = -11 + 6

b = -5

The parallel line equation is  y = 2x - 5.

Answer 3

2. Given point ( -3,0) and the equation is y = 6x+3

The parallel line equation for the given equation y = 6x+3 is y = 6x+k --------------(1)

Equation (1) passes through the point (-3,0)

Say (x,y) = (-3,0)

Substitute (x,y) = (-3,0) in (1)

i.e, 0 = 6(-3) + k

0 = -18 + k

Add 18 to each side

18 = k

Therefore k=18

Substitute k = 18 in (1)

y = 6x+18

Therefore slope intercept form is y = 6x + 18