Expert: Josh - 1/6/2004

Question
i'm having trouble with these type of problems:

solve and graph the solution set for each open expression:     ( , )=absolute value sign

2(x) -6- (x)less than or equal to 2(x) +10

also to graph :

5x-2y=10        x-y +10<0

Answer
Hi Elizabeth,

You need to remember two rules before we solve this problem.
---------------------------------------------------------
Rule 1: we DON'T have to worry about flipping the signs (from > to < or from >= to <= AND vice versa) when we ADD or SUBTRACT the same quantity from both sides.
Rule 2: we DO have to flip the sign if we MULTIPLY or DIVIDE both sides by a negative number.
---------------------------------------------------------
To see why this is so, consider the following example.
Example 1: Start with a true statement, say, 1 <= 2.

According to rule 1,
if we add the same number to both sides, it remains true without the need of flipping the sign.
For instance, 1+4 <= 2+4, i.e., 5 <=6
Same story with subtraction.
For instance, 1-4 <= 2-4, since -3<= -2.

But if we multiply both sides by -1 without changing the direction of the sign,
-1 < -2 is NOT TRUE.
We'll get it right by applying Rule 2,
-1 > -2 is TRUE.

Solution to your problem:
Step 1: simplify the inequality first.
We have 2|x|-6-|x|<=2|x|+10.
This is equivalent to |x|-6<=2|x|+10
Subtracting |x| from both sides, we get
-6<=|x|+10
Adding -10 to both sides,
-16<=|x|

Something wrong here, because |x| must be positive. Otherwise, we cannot interpret |x| as the distance measured from the origin x=0 (in the positive or negative sense).
Are you sure you copied the question right?

Q2. For 5x-2y=10, make y the subject. Put it to one side of the equation. ALways express it in the standard form, like
  y=mx+b  ...[#1]
Where m is the "slope" of the straight line, b is the y-axis intercept, when x=0.

5x-2y=10
5x-10=2y
y=(5/2)x-5 ...[#2]

Comparing [#1] and [#2], we see that "b"=-5, "m"=2.5
i.e., you graph a straight line, with an upward slope of 2.5 units rise over 1 unit run. It passes through y=-5, when x=0.

Q3. Do the same for x-y +10<0
x+10 < y ...[#3]
This describes an open region in the 2D plane.
Dotted line has gradient 1, y-intercept of y=10, when x=0.
To see which side of this dotted line is included in this region, pick an arbitrary point,
say, (x,y)=(0,0) for simplicity,
substitute back into inequality [#3] to see if it is true.
If it is true, this side is shaded in.
In this case, when x=0,y=0,
0+10 < 0 is FALSE.
So the side of the dotted line not containing (0,0) should be shaded in.