### Video Transcript

Solve the simultaneous equations π¦
equals π₯ minus two, π₯ minus two squared plus π¦ minus three squared is equal to
nine.

The first thing to notice is that
this is a linear-quadratic system of equations. The first equation π¦ equals π₯
minus two is a linear equation, as the highest power of π₯ and π¦ that appears is
one. And the second equation π₯ minus
two squared plus π¦ minus three squared equals nine is a quadratic equation because
once weβve distributed the parentheses, we will have both π₯ squared and π¦ squared
terms.

Weβre going to use the method of
substitution to answer this problem. Now, our first equation is π¦
equals π₯ minus two. And we notice that the expression
π₯ minus two appears in the second equation. So what we can do is replace π₯
minus two with π¦ in our second equation. Doing so gives π¦ squared plus π¦
minus three squared equals nine. And so we have an equation in π¦
only. Itβs a quadratic equation which we
can solve.

We begin by distributing the
parentheses in π¦ minus three squared. And we remember that π¦ minus three
all squared means π¦ minus three multiplied by π¦ minus three. So when we distribute the
parentheses, we have four terms. And these four terms then simplify
to π¦ squared minus six π¦ plus nine. Collecting like terms on the
left-hand side, we have the quadratic equation two π¦ squared minus six π¦ plus nine
equals nine.

Now, we notice that as we have
positive nine on each side, these two terms will directly cancel one another out,
which is equivalent to subtracting nine from each side of the equation. This leaves us with the simplified
equation two π¦ squared minus six π¦ equals zero, which we can solve by
factoring. The highest common factor of two π¦
squared and negative six π¦ is two π¦. And to make two π¦ squared, we have
to multiply two π¦ by π¦. And to make negative six π¦, we
have to multiply two π¦ by negative three. So our quadratic in its factored
form is two π¦ multiplied by π¦ minus three is equal to zero.

To solve, we take each factor in
turn, set it equal to zero, and then solve the resulting linear equation. The first equation is two π¦ equals
zero, which we can solve by dividing each side by two to find that π¦ is equal to
zero. The second equation is π¦ minus
three equals zero. We solve by adding three to each
side giving π¦ equals three. So we find that our solution to
these simultaneous equations includes two π¦-values, π¦ equals zero and π¦ equals
three.

We now need to find the
corresponding π₯-values. And to do this, we substitute each
π¦-value into the linear equation. It must be the linear equation into
which we substitute because itβs possible that, on the quadratic curve, there is
more than one point where π¦ equals zero or π¦ equals three. But on the linear graph, the
straight line, there will only be one point where π¦ equals zero and one point where
π¦ equals three. So we must substitute into the
linear equation.

When π¦ equals zero, we have zero
equals π₯ minus two. And adding two to each side, we
find that π₯ is equal to two. When π¦ equals three, we have that
three is equal to π₯ minus two. And adding two to each side, we
find that π₯ is equal to five. So the solution to this pair of
simultaneous equations is two pairs of π₯π¦-values. π₯ equals two and π¦ equals zero
and π₯ equals five and π¦ equals three.

Itβs important to understand that
these solutions come in pairs. And we canβt mix and match the π₯-
and π¦-values. For example, π₯ equals two and π¦
equals three is not a valid solution to this pair of simultaneous equations, which
you can confirm by substituting the values into the linear equation or, indeed, the
quadratic.

Remember, we used the method of
substitution to answer this question by replacing π₯ minus two with π¦. It would also have been possible to
replace π¦ in the second set of parentheses with π₯ minus two to give an equation in
π₯ only. In this case, weβd have found our
solutions for π₯ first and then substitute it back into the linear equation to find
our solutions for π¦. Our final answer is that π₯ equals
two and π¦ equals zero or π₯ equals five and π¦ equals three.