Give the X and Y intercepts of the equation, 20X + 16Y= 80

Lets find (X) Then (y) values accordingly:-

1) Finding (X) Value

20x + 16y = 80
Subtract both sides by (-16y)
(20x + 16y) + (-16y) = 80 + (-16y)
Remove +16y & (-16y) next to 20x
20x -16y + 80
10x/20 = -16y + 80/20
X = -2^4 * y + 2 ^4 * 5/20
X = 2^4 * [(2^4 * y/2^4)/2^4 + (2^4 * 5)/2^4]/20
X = 2^4 * (-y + 5)/20
X = 2^4 * (-y + 5)/2^2 * 5
X = 2^2 * (-y + 5)/5

X = 4 * (-y + 5)/5

2) Finding (Y) Value
20x + 16y = 80
(20x + 16y) + (-20x) = 80 - 20x)
20x + 16y - 20x = 80 - 20x
20x - 20x + 16y = -20x + 80
(20x - 20x = 0)
0 + 16y = -20x + 80
16y = -20x + 80
16y/16 = (-20x + 80)/16
Y = -(2^2 * 5)X + 2^4 * 5/16
Y = [2^2 * 5 * (2^2 * 5x/2^2 * 5 + 2^4 * 5/2^2 * 5)]/16
Y = [2^2 * 5 * (-x + 2^4-2)]/16
Y = [2^2 * 5 * (-x + 2^2)]/16
Y = [2^2 * 5 * (-x + 4)]/16
Y = 2^2 * 5 * (-x + 4)/2^4
Y = [5 * (-x + 4)]/2^4-2
Y = [5 * -x + 4)]/2^2

Y = [5 * (-x + 4)]/4

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True is the answer. hthhth