Graph the solution of each compound inequality
on a number line.
The solution given before this is incomplete
because you are supposed to combine them on
ONE number line:
y - 2 < 4 AND y + 4 > 7
Solve the first one: | Solve the second one:
|
y - 2 < 4 | y + 4 > 7
y < 6 | y > 3
So the simplified version is:
y < 6 AND y > 3
################################################
Draw three number lines, like this:
--------------------------------
-2 -1 0 1 2 3 4 5 6 7 8
--------------------------------
-2 -1 0 1 2 3 4 5 6 7 8
--------------------------------
-2 -1 0 1 2 3 4 5 6 7 8
################################################
Mark the numbers 3 and 6 on all three
number lines:
----------------o--------o------
-2 -1 0 1 2 3 4 5 6 7 8
----------------o--------o------
-2 -1 0 1 2 3 4 5 6 7 8
----------------o--------o------
-2 -1 0 1 2 3 4 5 6 7 8
################################################
Label the first number line " y < 6 "
Label the second number line " y > 3 "
Label the third number line " y < 6 AND y > 3 "
y < 6
----------------o--------o------
-2 -1 0 1 2 3 4 5 6 7 8
y > 3
----------------o--------o------
-2 -1 0 1 2 3 4 5 6 7 8
y < 6 AND y > 3
----------------o--------o------
-2 -1 0 1 2 3 4 5 6 7 8
################################################
Since the first one is y < 6, shade the part
left of 6
y < 6
<=========================o------
-2 -1 0 1 2 3 4 5 6 7 8
y > 3
----------------o--------o------
-2 -1 0 1 2 3 4 5 6 7 8
y < 6 AND y > 3
----------------o--------o------
-2 -1 0 1 2 3 4 5 6 7 8
################################################
Since the second one is y > 3, shade the part
right of 3:
y < 6
<=========================o------
-2 -1 0 1 2 3 4 5 6 7 8
y > 3
----------------o===============>
-2 -1 0 1 2 3 4 5 6 7 8
y < 6 AND y > 3
----------------o--------o------
-2 -1 0 1 2 3 4 5 6 7 8
################################################
Now to shade the last number line,
we will shade only the part where
both the first number line is shaded
AND the second one is shaded. Only
the part between 3 and 6 is shaded in
both the first number line AND the
second. So we just shade that part on
the third number line:
y < 6
<=========================o------
-2 -1 0 1 2 3 4 5 6 7 8
y > 3
----------------o===============>
-2 -1 0 1 2 3 4 5 6 7 8
y < 6 AND y > 3
----------------o========o------
-2 -1 0 1 2 3 4 5 6 7 8
That last number line is the final
answer. Sometimes teachers and
books tell you to put parentheses
if the end point is included and
brackets if it's not included.
Neither of these endpoints
is included, so the final answer in
that case would be like this:
y < 6 AND y > 3
----------------(========)------
-2 -1 0 1 2 3 4 5 6 7 8
Incidentally y > 3 can be rewritten
as 3 < y and so
3 < y AND y < 6
can be written all together as
3 < y < 6
Edwin