LESSON:
Systems of Linear Equations: Solving Equations
CONSISTS OF:
- Use substitution methods
- Use linear combination methods
NOTES:
- Answers are not in reduced
form, unless stated otherwise.
WORKPLACE:
ANS:
NULLx5 , 2 1/2 -5 1/4 , 2x + 1 , 3 , x , -5 , 2x - 3 , 6 , 3x , 2 , 1 , 5/3 , 10/3 , (-1/6)x , 20 , 11 2/3 , x - 5 , 4 , -3 , 2x - 3 , 18 , 1 , (-1/3)x + 2 , 2 , (-7/3)x , -6/7 , 16/7 , 8 - 2x , -2x + 8 , 6 , 4 , 5 - x , -4 , 9 , 3 + 2x , 0 , 3 , 4 , 1 , -4 , 7 , 7/2 , -8 , -16 , 8 , 13 , -4 , 4 , -9 , 3 , -4y , -6 , 6/5 , 44/5 , 11/5 , 21 , 30 , 1 , -15 , 1 , 1 , -21 , -16 , 3/5 , 1 - 3 , -4 - -9 , 2 - 1 , -1 - -3 , 2/3 - 1/3 , 15/7 - 10/7 , 0 - 2 , 2 - 5 , 6 - 10 ,
LESSON:
Linear Inequalities: Graphing Inequalities
CONSISTS OF:
- Write linear inequalities to
describe given graphs
- Graph linear inequalities
INFO:
Previously we have only seen inequalities on a number line:
x < 3
5 < x
-
x < -2
These inequalities involve only a single variable.
This lesson introduces inequalities involving TWO VARIABLES
These inequalities will be graphed in a plane.
For y=3, this graph shows x>5.
+
|
|
|
|
+
|
| o---->
|
|
----------------+----------------
- - - | - - -
|
|
|
+
|
|
|
|
+
Describe in your own words the graph of y<mx+b and the graph of y>mx+b. Then press ENTER.
The area below the line y=mx+b is the graph of y<mx+b. The area above the line y=mx+b is the graph of y>mx+b. Notice the from in which each inequality is written.
Before pressing ENTER, guess whether the graph of 2x+1<y is above or below the line 2x+1=y.
The shaded area above the line is the graph of 2x+1<y.
The inequality sign helps determine if the graph is above or below the line as long as the inequality is in the form y< or y> .
An additional method for deciding where the shade is by checking the points in the plane.
The point (3,9) satisfies the inequality 2x+1<y because:
2(3) + 1<9
To graph the inequality
y<(1/2)x+3
first graph the line
y=(1/2)x+3
then shade the area below the line.
WORKPLACE:
ANS:
>2 , <-2 , >2 , <4 , <0 , 5-7 , 4-7 , 3-7 , 3--1 , 4--1 , 5--1 , x>-2 , y>2 , y<=5 , x<-1 , y>=0 , YES , NO , YES , x=7 , LEFT , y=-1 , BELOW , y=-3 , ABOVE , x=-4 , RIGHT , < , > , < , < , > , YES , NO , YES , YES , NO , NO , YES , NO , y=2x-1 , ABOVE , y=-3x+2 , BELOW , y=(2/3)x+1 , BELOW , -x=y , ABOVE , -2>=x , y<5 , x>-3 , x+2<y , y<=(1/2)x , YES , NO , BELOW , ABOVE , ABOVE
LESSON:
Linear Inequalities: Systems of Inequalities
CONSISTS OF:
- Graph systems of inequalities
- Write compound algebraic
sentences to desribe
intersections or unions of
simple inequalities
INFO:
The point (5,7) satisfies the compund sentence:
x>3 AND y>5
because
5 is greater than 3 and
7 is greater than 5.
The point (1,7) does NOT satisfy the compound sentence:
x>3 AND y>5
because
1 is not greater than 3.
The point (5,2) does NOT satisfy the compound sentence:
x>3 AND y>5
because
2 is not greater than 5.
The point (2,4) does NOT satisfy the compound sentence:
x>3 AND y>5
because
2 is not greater than 3 and
4 is not greater than 5.
The point (0,-3) satisfies the compound sentence:
x>3 OR y<-2
because
-3 is less than -2.
The point (4,2) satisfies the compound sentence:
x>3 OR y<-2
because
4 is greater than 3.
The point (6,-6) satisfies the compound sentence:
x>3 OR y<-2
because
6 is greater than 3, and
-6 is less than -2.
The point (0,5) does NOT satisfy compound sentence:
x>3 OR y<-2
because
0 is not greater than 3 and
5 is not less than -2.
The inequality desribing this shaded area is x>5.
The inequality describing this shaded area is y>8.
The compound sentence describing the double shaded area is:
y>5 AND y>8
The compound sentence describing the ENTIRE area is:
x>5 OR y>8
Before pressing ENTER think of a compound that desribes the double shaded area on the graph.
The intersection that describes the double shaded area is:
x>6 AND y>3
Think of a compound sentence that describes the entire shaded area on the graph. Then press ENTER.
The union that describes the enter shaded area is:
x>-1 OR y<2
The equation of this line is y=(2/3)x+4.
The inequality describing the shaded area is y>(2/3)x+4.
The equation fo the second line is y=(4/5)x+10.
The inequality describing the second shaded area is Y<(-4/5)x+10.
The compound sentence describing the DOUBLE shaded area is:
y>(2/3)x+4
AND
y<(-4/5)x+10
The compound sentence describing the ENTIRE shaded area is:
y>(2/3)x+4
OR
y<(-4/5)x+10
---
The DOUBLE shaded area is the INTERSECTION of the two inequalities.
The ENTIRE shaded area is the UNION of the two inequalities.
WORKSPACE:
ANS:
YES , NO , NO , NO , YES , NO , YES , NO , NO , NO , YES , NO , YES , YES , YES , NO , YES , NO , YES , YES , NO , YES , YES , AND , AND , OR , OR , AND , AND , OR , (-1/2)x-1 , < , (3/4)x , < , and , 2x+4 , > , -x-2 , > , or , NO , YES , YES , NO , NO , YES , OR , x<-2 and y>3 , x>y and ... , y<5 or ...
LESSON:
Linear Inequalities: Absolute Value
CONSISTS OF:
- Solve linear inequalities
- Solve and graph equations and
inequalities involving absolute
value
INFO:
|4|
stands for "the absolute value of four."
|-4|
stands for "the absolute value of -4."
|5|
stands for "the absolute value of five."
|-10|
stands for "the absolute value of -10."
The absolute value of a number gives its distance from zero on a number line...
6 units
<------------
<|-|-|-|-|-|-|-|-|-|-|-|-|>
6 5 4 3 2 1 0 1 2 3 4 5 6
|-6| = 6
|5| = 5
|-4| = 4
|0| = 0
The absolute value of a number gives its distance from zero on a number line.
Before pressing ENTER, think of another way to describe absolute values of numbers.
The absolute value of a number always is POSITIVE: it gives the size of the number without regard to its sign (+ or -),
Two numbers satisfy this equation.
|x| = 7
They are 7 and -7.
Both are 7 units from 0 on a number line.
Using absolute values, the above two poinst can be described with a single equation:
|x| = 7
Both points are 7 units from 0 on a number line.
|11-4|
gives the distance between 11 and 4 on a number line. (7 units)
ANS:
11 , 8 , 7 , -11 , 11 , -8 , 8 , -15 , 15 , |x|=11 , |x|=8 , |x|=15 ,