Question

What is the equation of a line that is parallel to 4x+6y=10 and passes through (-4, 1)

Answer 1

Anybody

Answer 2

\[
4x + 6y = 10 \quad \to \quad 6y = -4x + 10 \\
y = \frac{-4}{6} + \frac{10}{6} = -\frac{2}{3}x + \frac{5}{3}
\]
by \(y = mx + h\) we get \(m = -\frac{2}{3}\).
This means that the slope of this straight line function is \(-\frac{2}{3}\).
Now, we need a function in form of \(y = -\frac{2}{3}x + C \) that would go through (-4,1) \\
So, we know x and y for our point, let's put it in to find C:
\[
x = -4 \quad \quad \quad y = 1 \\
y = -\frac{2}{3}x + C \quad \to \quad 1 = -\frac{2}{3}(-4) + C \\
1 = \frac{8}{3} + C \quad \to \quad \frac{3}{3} - \frac{8}{3} = C \\
C = -\frac{5}{3}
\]
So our function is
\[
y = -\frac{2}{3}x - \frac{5}{3}
\]
Small check:
\[
x = -4 \quad \implies \quad y = -\frac{2}{3}(-4) - \frac{5}{3} \\
y = \frac{8}{3} - \frac{5}{3} = \frac{3}{3} = 1
\]