i) The coefficients of the equation of the line are a = 20 / 3 and b = 160 / 21.
ii) The equation of the line in standard form is (20 / 3) · x + (160 / 21) · y = - 20.
iii) The x-intercept and y-intercept of the line are (- 3, 0) and (0, - 21 / 8).
iv) Two alternative solutions of the equation of the line are 20 · x + (160 / 7) · y = - 60 and 140 · x + 160 = 420.
How to derive the equation of a line?
In this problem we know that form of an equation of the line and two points, on which the line pass through. i) We determine the values of the coefficients a and b by solving the following system of linear equations:
5 · a - 7 · b = - 20
- 3 · a = - 20
Whose solution is a = 20 / 3 and b = 160 / 21.
ii) The equation of the line in standard form is (20 / 3) · x + (160 / 21) · y = - 20.
iii) Now we find the coordinates of the intercepts of the line:
x-Intercept
(20 / 3) · x = - 20
x = - 3
y-Intercept
(160 / 21) · y = - 20
y = - 21 / 8
iv) We can find two alternative solutions by using multiples:
20 · x + (160 / 7) · y = - 60
140 · x + 160 = 420
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