Question

Select the correct answer. Rewrite the following equation as a function of x.

Answer 1

Option B:

${f(x)}=9280-20x$

Solution:

$\frac{1}{16} x+\frac{1}{320} y-29=0$

$\frac{1}{16} x+\frac{1}{320} y-\frac{29}{1} =0$

Take LCM of the denominators and Make the denominators same.

LCM of 16, 320, 1 = 320

$\frac{1\times20}{16\times20} x+\frac{1}{320} y-\frac{29\times 320}{1\times 320} =0$

$\frac{20}{320} x+\frac{1}{320} y-\frac{9280}{320} =0$

All the denominators are same, so you can write in one fraction.

$\frac{20x+y-9280}{320}=0$

Do cross multiplication.

${20x+y-9280}=0\times 320$

${20x+y-9280}=0$

Add 9280 on both sides of the equation.

${20x+y}=9280$

Subtract 20x on both sides of the equation.

${y}=9280-20x$

Let y = f(x).

${f(x)}=9280-20x$

Hence Option B is the correct answer.