Figure 2.1a

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# Figure 2.1a - PowerPoint PPT Presentation

Technology 2.1. Evaluating Expressions. Figure 2.1a. For Figure 2.1a,. To evaluate an algebraic expression, we may substitute the value(s) in place of the variable(s) and evaluate the numeric expression. Be sure to enclose the values substituted in parentheses. (. 3. ). x 2. -. 4. (.

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Technology

2.1

Evaluating Expressions

Figure 2.1a

For Figure 2.1a,

To evaluate an algebraic expression, we may substitute the value(s)

in place of the variable(s) and evaluate the numeric expression. Be

sure to enclose the values substituted in parentheses.

(

3

)

x2

-

4

(

1

)

(

2

)

ENTER

(

(-)

5

)

x2

-

4

(

1

)

(

2

)

ENTER

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Evaluating Expressions

Technology

2.1

Figure 2.1b

For Figure 2.1b,

To evaluate an algebraic expression, we may store the value for the variables

and enter the algebraic expression. For ease in reading, we will enter all of these

commands as one entry. To do so, we separate each entry with a colon, .

Store the three values for a, b, and c separated by colons.

ALPHA

:

1

ALPHA

A

ALPHA

:

3

ALPHA

B

ALPHA

:

2

ALPHA

C

ALPHA

:

Enter the expression.

ALPHA

B

x2

-

4

ALPHA

A

ALPHA

C

ENTER

2 of 3

Technology

2.1

Evaluating Expressions

Figure 2.1b

In order to enter the second expression without retyping, recall

the previous entry and edit it. Press to recall the

previous entry. Then edit the previous entry, using the arrow

keys in combination with delete, , and insert .

2nd

ENTRY

2nd

INS

DEL

Move the cursor to the left, using the arrow keys.

Place the cursor on top of the 3 and delete 3, . Insert the

new value for b, -5,

2nd

ENTRY

DEL

2nd

INS

(-)

5

ENTER

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Technology

2.2

Testing Algebraic Equations

Determine whether x = 5 is a solution

of the equation 4x - 3 = 3x + 2.

For Figure 2.2a,

Figure 2.2a

Check the value given by evaluating the expression on the left and the

expression on the right. Store the given value for the variable, x = 5.

5

X,T,,n

Separate the entries.

ALPHA

:

Enter the right side and the left side separately.

4

X,T,,n

-

3

ENTER

Since 17 = 17, x = 5 is a solution.

3

X,T,,n

+

2

ENTER

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Technology

2.2

Testing Algebraic Equations

Determine whether x = 5 is a solution

of the equation 4x - 3 = 3x + 2.

For Figure 2.2b,

Figure 2.2b

Check the value by using the TEST function of the calculator.

Store the given value for the variable, x = 5.

5

X,T,,n

Separate the entries.

ALPHA

:

Enter the the equation. The equals sign is under TEST menu option 1.

4

X,T,,n

-

3

2nd

TEST

1

3

X,T,,n

+

2

ENTER

The calculator returns a 1 to indicate that the equation is true and returns a 0 to

indicate that the equation is false. Since the calculator returned a 1, x = 5 is a

solution of the equation.

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