Eliminate the absolute value
sign
by writing: x + 1 > 3 or x +
1 <
-3
Solve each sentence for x:
x > 2 or x < -4
Since this uses x,
x
> 2 is the area to the right of x = 2,
and
x
< - 4 is the area to the
left of -4.
Shade those areas. The union of these shaded areas is the answer.
Graph:
y < x3 +
2x2
Graph the polynomial y =
x3 + 2x2 first.
Since this inequality says y <
x3 + 2x2 , shade the area below the
graph
where the y-values
are small. The shaded area represents the
solution set of this inequality.
Graph the system:
y > -x
y < (x + 2)2
-4 < y < 6
Graph each inequality as though it were
an
equation. If y is greater than something, shade
up;
and, if y is less than something, shade
down.
The intersection of these shaded areas is the
proper graph. It's the part on the right side
between
the lines below the parabola and above
the line.
Graph: |x| - |y| < 4
Eliminate the absolute values by writing four
sentences.
Each sentence is obtained by using x and y values in a specific
quadrant.
Quadrant
I:
x - y < 4
Quadrant II:
-x - y < 4
Quadrant III:
-x + y < 4
Quadrant IV: x + y
< 4
Solve each one for y:
y > x - 4
y > -x - 4
y < x + 4
y < -x + 4
Graph each of these as though they were equalities,
shading
the appropriate area. The union of these areas is the proper
graph.
The answer is all the shaded areas put together.
Graph: |x| - |x - 2y| < 4
Rewrite this inequality without absolute
values: x - (x - 2y) < 4 when x
and x - 2y are both positive
-x - (x - 2y) < 4 when x is negative
and
x - 2y is positive
-x + (x - 2y) < 4 when x and x -
2y
are both negative
x + (x - 2y) < 4 when x is positive and
x - 2y is negative
Solve for y in each one and graph each equation, shading the appropriate area:
2y <
4
y <
2
y < 2
-2x + 2y < 4
2y < 2x +
4
y < x + 2
-2y <
4
y >
-2
y > -2
2x - 2y <
4
-2y < -2x +
4
y > x - 2
The union of these areas (all sets of ordered
pairs) is the solution set.