### Video Transcript

Find the π₯-coordinate of the point
at which the straight line three π₯ plus nine π¦ equals zero cuts the π₯-axis.

There are lots of ways of
approaching this question. Letβs begin by considering the
π₯π¦-plane as shown. Any point which cuts the π₯-axis
will have a π¦-coordinate equal to zero. This means that we can substitute
π¦ equals zero into the equation three π₯ plus nine π¦ equals zero. This gives us three π₯ plus nine
multiplied by zero equals zero.

As nine multiplied by zero is zero,
we are left with three π₯ is equal to zero. We can then divide both sides of
this equation by three. On the left-hand side the threes
cancel, and on the right-hand side zero divided by three is zero. The π₯-coordinate of the point at
which the straight line three π₯ plus nine π¦ equals zero cuts the π₯-axis is
zero.

An alternative method would be to
rewrite our equation in slopeβintercept form, π¦ equals ππ₯ plus π, where π is
the slope or gradient and π is the π¦-intercept. Subtracting three π₯ from both
sides of the original equation, we have nine π¦ is equal to negative three π₯. We can then divide both sides of
the equation by nine such that π¦ is equal to negative three-ninths π₯. As both the numerator and
denominator of the fraction are divisible by three, this can be rewritten as π¦ is
equal to negative one-third π₯.

The equation three π₯ plus nine π¦
equals zero has a slope or gradient equal to negative one-third and a π¦-intercept
equal to zero. This linear equation can be drawn
on the π₯π¦-plane as shown. As this passes through the origin,
this confirms that the π₯-coordinate where the line cuts the π₯-axis is zero.