Option B:
[tex]${f(x)}=9280-20x[/tex]
Solution:
[tex]$\frac{1}{16} x+\frac{1}{320} y-29=0[/tex]
[tex]$\frac{1}{16} x+\frac{1}{320} y-\frac{29}{1} =0[/tex]
Take LCM of the denominators and Make the denominators same.
LCM of 16, 320, 1 = 320
[tex]$\frac{1\times20}{16\times20} x+\frac{1}{320} y-\frac{29\times 320}{1\times 320} =0[/tex]
[tex]$\frac{20}{320} x+\frac{1}{320} y-\frac{9280}{320} =0[/tex]
All the denominators are same, so you can write in one fraction.
[tex]$\frac{20x+y-9280}{320}=0[/tex]
Do cross multiplication.
[tex]${20x+y-9280}=0\times 320[/tex]
[tex]${20x+y-9280}=0[/tex]
Add 9280 on both sides of the equation.
[tex]${20x+y}=9280[/tex]
Subtract 20x on both sides of the equation.
[tex]${y}=9280-20x[/tex]
Let y = f(x).
[tex]${f(x)}=9280-20x[/tex]
Hence Option B is the correct answer.
