2.2 variables, evaluation and expressions w

transcript

Variables, Evaluation and Linear Expressions

http://www.lahc.edu/math/frankma.htm

In mathematics we use symbols such as x, y and z to represent numbers.

These symbols are called variables because their values change according to the values assigned to them.

These symbols are called variables because their values change according to the values assigned to them. We use numbers, variables and mathematics operations to form (variable)-expressions to describe calculation procedures.

For example, if x represents the number of apples and each apple cost $2,

For example, if x represents the number of apples and each apple cost $2, then “2*x” or “2x” is the expression for the cost of x apples.

For example, if x represents the number of apples and each apple cost $2, then “2*x” or “2x” is the expression for the cost of x apples. So if we have 6 apples,

For example, if x represents the number of apples and each apple cost $2, then “2*x” or “2x” is the expression for the cost of x apples. So if we have 6 apples, by setting the x as (6) in the expression 2x,

= $12 for the cost of 6 apples. we obtain 2(6)

The value “6” for x is called the input (value).

$2the input x = 6:

The answer “$12” is called the output (value).

$2the input x = 6: the output

2(6) = $12$2 $2

The above process of replacing the variable(s) with input value(s) to find the output is called evaluation.

The answer “$12” is called the output (value).

$2the input x = 6: the output

2(6) = $12$2 $2

Example A. Evaluate the following expressions with the given input values. Specify the input and the output.

a. 3 + x with x = 5

b. 5 + 3y with y = 2

c. 5z2 with z = 3

a. 3 + x with x = 5

b. 5 + 3y with y = 2

c. 5z2 with z = 3

To evaluate an expression, replace the variable(s) with the assigned input-value(s) enclosed by ( )’s, then compute the resulting expression following the rules of order of operations.

a. 3 + x with x = 5

b. 5 + 3y with y = 2

c. 5z2 with z = 3

by (5)replace x

a. 3 + x with x = 5Set x = 5 we’ve 3 + x = 3 + (5)

b. 5 + 3y with y = 2

c. 5z2 with z = 3

by (5)replace x

b. 5 + 3y with y = 2

c. 5z2 with z = 3

by (5)replace x= 8.

b. 5 + 3y with y = 2

c. 5z2 with z = 3

The input is x = 5 and the output is 8.

by (5)replace x= 8.

b. 5 + 3y with y = 2

c. 5z2 with z = 3

by (5)replace x= 8.

5 + 3y

by (2)replace y

b. 5 + 3y with y = 2Set y = 2 we’ve 5 + 3(2)

c. 5z2 with z = 3

by (5)replace x= 8.

5 + 3y

by (2)replace y

c. 5z2 with z = 3

by (5)replace x= 8.

5 + 3y

by (2)replace y

= 5 + 6 = 11.

c. 5z2 with z = 3

The input is y = 2 and the output is 11.

by (5)replace x= 8.

5 + 3y

by (2)replace y

= 5 + 6 = 11.

c. 5z2 with z = 3

by (5)replace x= 8.

5 + 3y

by (2)replace y

= 5 + 6 = 11.

by (3)replace z

c. 5z2 with z = 3Set z = 3 we’ve 5(3)2

by (5)replace x= 8.

5 + 3y

by (2)replace y

= 5 + 6 = 11.

by (3)replace z

= 5 * 9 = 45

by (5)replace x= 8.

5 + 3y

by (2)replace y

= 5 + 6 = 11.

by (3)replace z

= 5 * 9 = 45

The input is z = 3 and the output is 45.

by (5)replace x= 8.

5 + 3y

by (2)replace y

= 5 + 6 = 11.

by (3)replace z

d. 5 + 3(20 – 3z2) with z = 2

by (2)replace z

5 + 3(20 – 3z2)

d. 5 + 3(20 – 3z2) with z = 2

Set z = 2 we’ve 5 + 3(20 – 3(2)2)

by (2)replace z

5 + 3(20 – 3z2)

d. 5 + 3(20 – 3z2) with z = 2

Set z = 2 we’ve 5 + 3(20 – 3(2)2) = 5 + 3(20 – 3*4)

by (2)replace z

5 + 3(20 – 3z2)

d. 5 + 3(20 – 3z2) with z = 2

Set z = 2 we’ve 5 + 3(20 – 3(2)2)

= 5 + 3(20 – 12)= 5 + 3(20 – 3*4)

by (2)replace z

5 + 3(20 – 3z2)

d. 5 + 3(20 – 3z2) with z = 2

Set z = 2 we’ve 5 + 3(20 – 3(2)2)

= 5 + 3(20 – 12)

= 5 + 3(8)

= 5 + 3(20 – 3*4)by (2)replace z

5 + 3(20 – 3z2)

d. 5 + 3(20 – 3z2) with z = 2

Set z = 2 we’ve 5 + 3(20 – 3(2)2)

= 5 + 3(20 – 12)

= 5 + 3(8)

= 5 + 3(20 – 3*4)

= 5 + 24 = 29

by (2)replace z

5 + 3(20 – 3z2)

d. 5 + 3(20 – 3z2) with z = 2

Set z = 2 we’ve 5 + 3(20 – 3(2)2)

= 5 + 3(20 – 12)

= 5 + 3(8)

= 5 + 3(20 – 3*4)

The input is z = 2 and the output is 29.= 5 + 24 = 29

by (2)replace z

5 + 3(20 – 3z2)

d. 5 + 3(20 – 3z2) with z = 2

Set z = 2 we’ve 5 + 3(20 – 3(2)2)

= 5 + 3(20 – 12)

= 5 + 3(8)

= 5 + 3(20 – 3*4)

by (2)replace z

5 + 3(20 – 3z2)

The word “evaluation” here does not mean “to judge” as in “student-evaluation”.

d. 5 + 3(20 – 3z2) with z = 2

Set z = 2 we’ve 5 + 3(20 – 3(2)2)

= 5 + 3(20 – 12)

= 5 + 3(8)

= 5 + 3(20 – 3*4)

by (2)replace z

5 + 3(20 – 3z2)

In mathematics, “evaluation” is the forward-computation where we plug in the known quantities in a formula to obtain the corresponding output.

d. 5 + 3(20 – 3z2) with z = 2

Set z = 2 we’ve 5 + 3(20 – 3(2)2)

= 5 + 3(20 – 12)

= 5 + 3(8)

= 5 + 3(20 – 3*4)

by (2)replace z

5 + 3(20 – 3z2)

Linear Expressions

d. 5 + 3(20 – 3z2) with z = 2

Set z = 2 we’ve 5 + 3(20 – 3(2)2)

= 5 + 3(20 – 12)

= 5 + 3(8)

= 5 + 3(20 – 3*4)

by (2)replace z

5 + 3(20 – 3z2)

Linear ExpressionsIn life, we often need to compute in the following manner:

d. 5 + 3(20 – 3z2) with z = 2

Set z = 2 we’ve 5 + 3(20 – 3(2)2)

= 5 + 3(20 – 12)

= 5 + 3(8)

= 5 + 3(20 – 3*4)

by (2)replace z

5 + 3(20 – 3z2)

Linear ExpressionsIn life, we often need to compute in the following manner: * perform a multiplication/division to obtain a product/quotient,

d. 5 + 3(20 – 3z2) with z = 2

Set z = 2 we’ve 5 + 3(20 – 3(2)2)

= 5 + 3(20 – 12)

= 5 + 3(8)

= 5 + 3(20 – 3*4)

by (2)replace z

5 + 3(20 – 3z2)

Linear ExpressionsIn life, we often need to compute in the following manner: * perform a multiplication/division to obtain a product/quotient,* then add/subtract another number with the above result.

Example B. a. An online order sells DVD’s for $10/disc with a $7 shipping and handling fee for each order. We ordered 6 DVD’s, how much is the total cost for the order?

Example B. a. An online order sells DVD’s for $10/disc with a $7 shipping and handling fee for each order. We ordered 6 DVD’s, how much is the total cost for the order? Calculate the cost of the 6 DVD’s,

10(6) + 7 = then add the $7 handling fee:

60 + 7 = $67.10(6) + 7 = then add the $7 handling fee:

So the total cost for the order is $67 for 6 DVD’s. 60 + 7 = $67.10(6) + 7 =

then add the $7 handling fee:

b. We change the order to 9 DVD’s, how much is the total cost?

10(9) + 7 = Multiply, then add:

90 + 7 = $97.10(9) + 7 = So the total cost for the order is $97 for 9 DVD’s. Multiply, then add:

c. Describe the input and output variables and what are the specific input and the output for each problem?

The input variable is “the number of DVD’s ordered”.

Calculate the cost of the 6 DVD’s,

90 + 7 = $97.10(9) + 7 = So the total cost for the order is $97 for 9 DVD’s.

The output variable is “the total cost”.

Multiply, then add:

then for the input x = 6, the output is $67,

Let’s name “the number of DVD’s ordered” as x, The output variable is “the total cost”.

Multiply, then add:

then for the input x = 6, the output is $67,

Let’s name “the number of DVD’s ordered” as x,

and for the input x = 9, the output is $97.

The output variable is “the total cost”.

Multiply, then add:

d. Write the expression for calculating the total cost of an order using the variable “x” where x = the number of DVD’s ordered,

The cost of the x DVD’s is 10x,

The cost of the x DVD’s is 10x, we have the expression for the total cost to be 10x + 7.

adding the $7 handling fee,

Example C. We have a filled 1,000-gallon water tank. On the average, we use 15 gallons of water each day. How much water is left in the tank after 10 days?How about after 1 month, i.e. 30 days? Write down the expression that calculates the amount of water left after x days.

Example C. We have a filled 1,000-gallon water tank. On the average, we use 15 gallons of water each day. How much water is left in the tank after 10 days?How about after 1 month, i.e. 30 days?

After 10 days, we used 15(10) = 150 gallons of water.

Write down the expression that calculates the amount of water left after x days.

After 10 days, we used 15(10) = 150 gallons of water.So there is 1,000 – 150 = 850 gallons left.

After 30 days, we used 15(30) = 450 gallons of water.

After x days, we used 15(x) = 15x gallons of water.So there is 1,000 – 15x gallons left.

Expressions of the form ax + b, where a and b are constants, i.e. fixed numbers like the $10/DVD and $7 fee in the last example, with x as the variable, are called linear expressions.

Variables, Evaluation and Linear ExpressionsLinear Expressions

Linear expressions do not involve x2, x3, etc.. or division by x so expressions such as 3x2 + 1 or 12 ÷ x are non-linear.

Linear expressions do not involve x2, x3, etc.. or division by x so expressions such as 3x2 + 1 or 12 ÷ x are non-linear.Linear expressions arise frequently in everyday computation. Because linear expressions are easy to compute, they also are utilized to estimate more complicated non-linear expressions.

2.2 variables, evaluation and expressions w

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