### Video Transcript

Find the three inequalities that
define the shaded region.

In order to find the inequalities
that define this region, we must first identify the equations of each of the given
lines. Probably, the easiest line to
identify is the horizontal line. This line passes through the
π¦-axis at negative three. In fact, every coordinate on this
line has a π¦-value of negative three. This means it has the equation of
π¦ is equal to negative three.

We can see that the shaded region
sits above the line π¦ is equal to negative three. It is also a solid line. Therefore, this inequality must be
π¦ is greater than or equal to negative three. Remember a dotted line is just
greater than or less than, whereas the solid line is greater than or equal to or
less than or equal to.

The other two equations are a
little trickier. Remember the equation of any
straight line is given as π¦ is equal to ππ₯ plus π, where π is the gradient and
π is the π¦-intercept. If we begin by looking at the
slightly steeper line, we can see that it has a π¦-intercept of positive three. The gradient is a little
trickier. The gradient is sometimes called
rise over run or change in π¦ over change in π₯.

If we choose two coordinates on
this line as shown, we can construct a right-angled triangle. The change in π¦ is given by the
height of the triangle which is positive four. The change in π₯ is the width of
the triangle, positive one. Its gradient is, therefore, four
divided by one which is four. The equation of this line is,
therefore, π¦ is equal to four π₯ plus three.

The shaded region lies to the left
of this line. In order to decide which way we
want the inequality symbol to go, we choose a coordinate on this left side of the
line. Letβs choose negative two and
negative three. In this coordinate, π₯ has a value
of negative two and π¦ has a value of negative three. So we can substitute these values
into the equation for our straight line. This gives us negative three on the
left-hand side of the equation and four multiplied by negative two plus three on the
right. That gives us a value of negative
five.

Negative three is greater than
negative five and the line is a solid line. So our inequality becomes π¦ is
greater than or equal to four π₯ plus three. Our final line clearly has a
π¦-intercept of negative one. Letβs construct another
right-angled triangle to help us find the gradient.

By marking two of the coordinates
we can choose, we can see the change in π¦ is the height of this triangle β itβs
positive two. The change in π₯ is the width of
the triangle which is four. The gradient of this line is,
therefore, two divided by four which is a half. The equation of this straight line
is, therefore, π¦ is equal to a half π₯ minus one. This time the shaded region lies
underneath or to the right of the line. So we choose a point on this
side. We can choose the same one as last
time β negative two and negative three.

Substituting these values in, we
get negative three on the left-hand side of our equation and a half multiplied by
negative two minus one on the right. That gives us negative two. Negative three is less than
negative two and itβs a solid line. So our inequality becomes π¦ is
less than or equal to a half π₯ minus one.

The three inequalities that define
the shaded region are π¦ is greater than or equal to negative three, π¦ is greater
than or equal to four π₯ plus three, and π¦ is less than or equal to a half π₯ minus
one.