we know that
The equation of the line in the slope-intercept form is equal to
\(y=mx+b\)
we have
\(x+4y=12\)
Solve for y
\(x+4y=12\)
Subtract \(x\) both sides
\(x+4y-x=12-x\)
\(4y=12-x\)
Divide by \(4\) both sides
\(4y/4=(12-x)/4\)
\(y=(12-x)/4\)
\(y=-\frac{1}{4}x+3\)
therefore
the answer is
\(y=-\frac{1}{4}x+3\)
