b The inequality will have the same boundaries as the one in Part A.
C
c By changing the inequality to an equation we can find the boundary point(s) of the inequality.
D
d The inequality will have the same boundaries as the one in Part C.
E
e By changing the inequality to an equation we can find the boundary point(s) of the inequality.
F
f The inequality will have the same boundaries as the one in part E.
G
g Examine the solution sets of the exercise pairs. Can you find a pattern?
A
aDiagram:
B
bDiagram:
C
cDiagram:
D
dDiagram:
E
eDiagram:
F
fDiagram:
G
g See solution.
Practice makes perfect
a To solve the inequality we should first find its boundary point(s). To do that, we will treat the inequality as an equality.
Inequality:& |2x+3|<5
Equality:& |2x+3|=5
This equation means that the distance is 5, either in the positive direction or the negative direction.
|2x+3|= 5 ⇒ l2x+3= 5 2x+3= - 5
To find the solutions to the absolute value equation, we need to solve both of these cases for x.
lc2x+3=5 & (I) 2x+3=-5 & (II)
(I), (II): LHS-3=RHS-3
l2x=2 2x=-8
(I), (II): .LHS /2.=.RHS /2.
lx_1=1 x_2=-4
The boundary points are located at x=1 and x=-4. Let's mark them on a number line. Since the inequality is strict, the boundaries are not included in the solution set. To determine where we should shade the number line, we will also include three test points.
Now, substitute x in the inequality with these values and see which one holds.
|c|c|c|
[-0.8em]
x & |2x+3|<5 & Evaluate [0.5em]
[-1em]
-6 & |2( -6)+3|? <5 & 9 ≮ 5 * [0.5em]
[-1em]
0 & |2( 0)+3|? <5 & 3 < 5 âś“ [0.5em]
[-1em]
2 & |2( 2)+3|? <5 & 7 ≮ 5 * [0.5em]
The inequality is true between the boundary points. Therefore, we should shade this region.
b Like in Part A, we will solve the inequality by first finding its boundary points. To do that, we will treat the inequality as an equality.
Inequality:& |2x+3|≥ 5
Equality:& |2x+3|=5
From Part A, we know that the solution to this equation is x=-4 and x=1. Let's substitute the same test points from Part A to see which ones makes the inequality true.
|c|c|c|
[-0.8em]
x & |2x+3|≥ 5 & Evaluate [0.5em]
[-1em]
-6 & |2( -6)+3|? ≥5 & 9 ≥ 5 ✓ [0.5em]
[-1em]
0 & |2( 0)+3|? ≥5 & 3 ≱ 5 * [0.5em]
[-1em]
2 & |2( 2)+3|? ≥5 & 7 ≥ 5 ✓ [0.5em]
The inequality is true to the left and to the right of the boundary points. Therefore, we should shade these regions. Notice that the inequality is non-strict which means the endpoints are filled.
c Like in previous parts, we will find its boundary points by treating the inequality as an equation.
Inequality:& |2x-3|< 5
Equality:& |2x-3|=5
This equation means that the distance is 5, either in the positive direction or in the negative direction.
|2x-3|= 5 ⇒ l2x-3= 5 2x-3= - 5
The solutions to the equations will give us the boundary points
lc2x-3=5 & (I) 2x-3=-5 & (II)
(I), (II): LHS+3=RHS+3
l2x=8 2x=-2
(I), (II): .LHS /2.=.RHS /2.
lx_1=4 x_2=-1
The boundaries are located at x=-1 and x=4. Let's mark them on a number line. Since the inequality is strict, the boundaries are not included in the solution set. To determine where we should shade the number line, we will also include three test points.
Now, substitute x in the inequality with these values to see which one makes the inequality true.
|c|c|c|
[-0.8em]
x & |2x-3|<5 & Evaluate [0.5em]
[-1em]
-2 & |2( -2)-3|? <5 & 7 ≮ 5 * [0.5em]
[-1em]
0 & |2( 0)-3|? <5 & 3 < 5 âś“ [0.5em]
[-1em]
5 & |2( 5)-3|? <5 & 7 ≮ 5 * [0.5em]
The inequality is true between the boundary points. Therefore, we should shade this region.
d Like in Previous Part, we will solve the inequality by first finding its boundary points. To do that, we will treat the inequality as an equality.
Inequality:& |2x-3|≥ 5
Equality:& |2x-3|=5
From Part C, we know that the solution to this equation is x=4 and x=- 1. Let's substitute these into the inequality to determine which ones holds true.
|c|c|c|
[-0.8em]
x & |2x-3|≥ 5 & Evaluate [0.5em]
[-1em]
-2 & |2( -2)-3|? ≥5 & 7 ≥ 5 ✓ [0.5em]
[-1em]
0 & |2( 0)-3|? ≥5 & 3 ≱ 5 * [0.5em]
[-1em]
5 & |2( 5)-3|? ≥5 & 7 ≥ 5 ✓ [0.5em]
The inequality is true to the left and to the right of the boundary points. Therefore, we should shade these regions. Notice that the inequality is non-strict which means the endpoints are filled.
e Like in previous parts, we should first find its boundary point(s). To do that, we will treat the inequality as an equality.
Inequality:& |3-2x|<5
Equality:& |3-2x|=5
This equation means that the distance is 5, either in the positive direction or the negative direction.
|3-2x|= 5 ⇒ l3-2x= 5 3-2x= - 5
To find the solutions to the absolute value equation, we need to solve both of these cases for x.
lc3-2x=5 & (I) 3-2x=-5 & (II)
(I), (II): LHS-3=RHS-3
l-2x=2 -2x=-8
(I), (II): .LHS /(-2).=.RHS /(-2).
lx_1=-1 x_2=4
The boundary points are located at x=-1 and x=4. Let's mark them on a number line. Since the inequality is strict, the boundaries are not included in the solution set. To determine where we should shade the number line, we will also include three test points.
Now, substitute x in the inequality with these values and see which one makes the inequality hold true.
|c|c|c|
[-0.8em]
x & |3-2x|<5 & Evaluate [0.5em]
[-1em]
-2 & |3-2( -2)|? <5 & 7 ≮ 5 * [0.5em]
[-1em]
0 & |3-2( 0)|? <5 & 3 < 5 âś“ [0.5em]
[-1em]
6 & |3-2( 6)|? <5 & 9 ≮ 5 * [0.5em]
The inequality is true between the boundary points. Therefore, we should shade this region.
f Like in Previous Part, we will solve the inequality by first finding its boundary points. To do that, we will treat the inequality as an equality.
Inequality:& |3-2x|≥ 5
Equality:& |3-2x|=5
From Part E, we know that the solution to this equation is x=-1 and x=4. Let's substitute these into the inequality to determine which ones holds true.
|c|c|c|
[-0.8em]
x & |3-2x|≥ 5 & Evaluate [0.5em]
[-1em]
-2 & |3-2( -2)|? ≥5 & 7 ≥ 5 ✓ [0.5em]
[-1em]
0 & |3-2( 0)|? ≥5 & 3 ≱ 5 * [0.5em]
[-1em]
6 & |3-2( 6)|? ≥5 & 9 ≥ 5 ✓ [0.5em]
The inequality is true to the left and to the right of the boundary points. Therefore, we should shade these regions. Notice that the inequality is non-strict which means the endpoints are filled.
g Examining the problems, we see that Part A, C and E are inequalities describing an expression that is less than 5. The remaining three Parts, B, D and F describes the same expressions but for values that are greater than or equal to 5.
2
a.& |2x+3|<5 &b.& |2x+3|≥5 [0.2em]
[-1em]
c.& |2x-3|<5 &d.& |2x-3|≥5 [0.2em]
[-1em]
e.& |3-2x|<5 &f.& |3-2x|≥5
The expressions in the left column represent all x-values that make the expressions less than 5. Together with the right column that represents all x-values that make the expressions greater or equal to 5, they cover the whole number line.
