Answer:
1) The equation of the line in slope-intercept form is
. The equation of the line in standard form is
.
2) The equation of the line in slope-intercept form is
. The equation of the line in standard form is
.
3) The equation of the line in slope-intercept form is
. The equation of the line in standard form is
.
4) The equation of the line in slope-intercept form is
. The equation of the line in standard form is
.
5) The equation of the line in slope-intercept form is
. The equation of the line in standard from is
.
Step-by-step explanation:
1) We begin with the slope-intercept form and substitute all known values and calculate the y-intercept: (
,
,
)



The equation of the line in slope-intercept form is
.
Then, we obtain the standard form by algebraic handling:

The equation of the line in standard form is
.
2) We begin with a system of linear equations based on the slope-intercept form: (
,
,
,
)
(Eq. 1)
(Eq. 2)
From (Eq. 1), we find that:

And by substituting on (Eq. 2), we conclude that slope of the equation of the line is:



And from (Eq. 1) we find that the y-Intercept is:



The equation of the line in slope-intercept form is
.
Then, we obtain the standard form by algebraic handling:


The equation of the line in standard form is
.
3) By using the slope-intercept form, we obtain the equation of the line by direct substitution: (
,
)

The equation of the line in slope-intercept form is
.
Then, we obtain the standard form by algebraic handling:

The equation of the line in standard form is
.
4) We begin with a system of linear equations based on the slope-intercept form: (
,
,
,
)
(Eq. 3)
(Eq. 4)
By applying (Eq. 4) on (Eq. 3), we find that the slope of the equation of the line is:



The equation of the line in slope-intercept form is
.
Then, we obtain the standard form by algebraic handling:

The equation of the line in standard form is
.
5) We begin with a system of linear equations based on the slope-intercept form: (
,
,
,
)
(Eq. 5)
(Eq. 6)
From (Eq. 5), we find that:

And by substituting on (Eq. 6), we conclude that slope of the equation of the line is:



And from (Eq. 5) we find that the y-Intercept is:


The equation of the line in slope-intercept form is
.
Then, we obtain the standard form by algebraic handling:


The equation of the line in standard from is
.