Solve the system of inequalities by
shading the correct region on the grid.
We’ve been given three inequalities
and asked to represent the solution to this system of inequalities on the coordinate
grid. To do so, we first need to draw
three lines on the grid, the lines with equations 𝑦 equals negative two, 𝑥 equals
negative two, and 𝑦 is equal to negative two 𝑥 plus three.
Let’s consider what each of these
lines look like. An equation of the form 𝑦 equals
constant is a horizontal line. Remember, when we’re representing
inequalities on coordinate grids, the type of line we draw depends on the type of
inequality. For strict inequalities, we use a
dotted line. And for weak inequalities, we use a
As our first inequality is a weak
inequality, we draw the line 𝑦 equals negative two as a solid line. As we’re looking for where 𝑦 is
greater than or equal to negative two, this means that we’re going to be shading
above the line. As we’ve drawn the line as a solid
line, points on the line will also be included in our solution. Lines with the equation 𝑥 is equal
to a constant are vertical lines.
As the second inequality is a weak
inequality, we draw the line 𝑥 equals negative two as a dotted line. The inequality is 𝑥 is greater
than negative two, which means we’re going to be shading points to the right of the
Let’s consider what the final line
looks like. If we compare this to 𝑦 equals
𝑚𝑥 plus 𝑐, the general form of the equation of a straight line, then we can see
that the 𝑦-intercept of this line is positive three. The gradient of this line is
negative two, which means, for every one unit we move to the right, the line moves
two units down.
As this inequality is a weak
inequality, we again draw a dotted line for the line 𝑦 is equal to negative two 𝑥
plus three. Now we need to consider which side
of this line we want to shade. To do so, let’s consider any point
that doesn’t lie on the line. So I’m going to consider the
origin, the point with coordinates zero, zero.
Let’s substitute zero for 𝑥 and
zero for 𝑦 into the equation of the line. This gives zero on the left-hand
side and negative two multiplied by zero plus three on the right-hand side. The right-hand side just simplifies
to three. What’s the relationship between
zero and three? Well, it’s that zero is less than
three, which means that this point does satisfy the inequality 𝑦 is less than
negative two 𝑥 plus three. And therefore, this point is in the
solution region that we’re looking for. This means that we want to shade
points that are to the left of the diagonal line we’ve drawn.
Putting all of this information
together for the three lines, so that’s shading points above 𝑦 equals negative two
to the right of 𝑥 equals negative two and below the diagonal line, we see that the
region that solves this system of inequalities is the triangular region here.
We’ve already tested that the third
inequality does hold true for the origin, which is a point in our region. But if you wanted to be extra
confident in your answer, you could pick another random point inside the region, for
example, the point with coordinates negative one, two, and confirm that the three
inequalities are indeed all true for this point.