Algebra 1

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# Algebra 1 - PowerPoint PPT Presentation

Algebra 1. Chapter 3.7 Formulas &amp; Functions. Objective. Students will solve formula’s for one of its variables and rewrite equations in function form. Before we begin…. In this lesson we will look at equations that have more than one variable…

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### Algebra 1

Chapter 3.7 Formulas & Functions

Objective
• Students will solve formula’s for one of its variables and rewrite equations in function form.
Before we begin…
• In this lesson we will look at equations that have more than one variable…
• More specifically, we will solve a formula for one of it’s variables and rewrite or transform an equation in function form…
• The key to being successful here is your ability to analyze an equation and make decisions that will help you solve the problem…
Solving Formulas
• A formula is an algebraic equation that relates two or more real-life quantities.
• A formula usually contains 2 or more variables…
• You should already be familiar with some formulas like perimeter, area, volume and surface area.
• In this lesson we will be transforming (changing) the formula to isolate the unknown variable…
• Let’s look at an example…
Example #1
• Using the formula for the area of a rectangle A = ℓw.
• Find ℓ in terms of A and w
• Use the new formula to find the length of a rectangle that has an area of 35 square feet and a width of 7 feet

The first step is to understand what the formula is saying. In this instance it says Area = length ● width

When you are asked to find ℓ in terms of A and w, you are being asked to isolate ℓ to one side of the equation

To do that, you undo the operations in the formula using just the variables…

Example #1 (continued)

A = ℓw

In this instance we will begin by working on the right side of the equation.

Since ℓw means to multiply, we will undo the multiplication by dividing both sides by w. Which will look like this

The w’s on the right side cancel out leaving ℓ

A = ℓw

w

w

This is the transformed formula with ℓ isolated on the right side

A = ℓ

w

Example #1 (continued)

A = ℓ

w

Now that we have the transformed formula all we have to do is substitute the given data into the formula and simplify to determine the length of the rectangle.

Use the new formula to find the length of a rectangle that has an area of 35 square feet and a width of 7 feet

35 = ℓ

7

5 = ℓ

• Of course, in the first example we used a simple formula so that you understand what this strategy is trying to accomplish.
• You can use this strategy for any formula…This strategy also has application in your science class with science formula’s
• Let’s look at a more complex formula…
Example #2

In this example we will use the formula for converting temperature from Celsius to Fahrenheit and solve for F

The first step is deciding which side of the equation to work on. Since F is on the right side we will begin there…

Second, I notice that the formula illustrates the distributive property…

I can do one of 2 things here….I can distribute the 5/9 or I can multiply by the reciprocal of 5/9 to get rid of the fraction on the right side of the equation…(we talked about this in a previous lesson…)

My decision is to multiply by the reciprocal to get rid of the fraction on the right side…

Example #2 (continued)

Original formula

Multiply by the reciprocal of 5/9. The fractions on the right side cancel out.

To undo the -32 on the right add 32 to both sides of the equation

The transformed formula with F isolated on one side of the equation

• In the previous example you were not given any data…all you were asked to do was solve for F, which means to isolate F on one side of the formula…
• Let’s look at another example where you have to transform the formula and use data to solve the problem…
• In this example we will use the formula for simple interest I=prt, where I = interest; p=principal; r=rate; t=time.
Example #3

I = prt

• Solve for r
• Find the interest rate (r) for an investment of \$1500 (p) that earned \$54 (I) in 1 year (t)

A: Transform the formula so that r is isolated to one side of the equation

I = prt

To do that you have to know that prt means multiply p times r times t. To undo multiplication use division like this:

Original Formula

I = prt

The p and the t on the right side cancel out leaving r

pt p t

I = r

The transformed formula r is isolated on one side of the equation.

pt

Example #3 (continued)

B. Find the interest rate (r) for an investment of \$1500 (p) that earned \$54 (I) in 1 year (t)

I = r

The transformed formula

pt

54 = r

Substitute the data into the formula

1500(1)

54 = r

Simplify

1500

0.036 or 3.6% = r

Solution

Equations in Function Form
• Continuing with the theme of isolating variables…
• A two-variable equation is written in Function Form if one of its variables is isolated to one side of the equation
• The isolated variable is called the output and it is a function of the other variable called the input.
• Let’s look at an example…
Example #4

Rewrite the equation 3x + y = 4 so that y is a function of x

In this example you are being asked to isolate y to one side of the equation. Once you do that the equation will be in function form

To transform this equation we will continue with the concepts that we have been working with, which is inverse operations

The first step is to decide which side of the equation you will be working on….In this case since y is on the left side I will begin there

On the left side is a +3x. To undo the +3x subtract 3x from both sides of the equation like this:

3x + y = 4

Since -3x and 4 are not like terms you cannot combine them

The 3x’s on the left cancel out leaving y

-3x -3x

y = - 3x + 4

Example #4 (continued)

This is the transformed equation:

y = - 3x + 4

This equation represents y as a function of x

What that means is that the value of y (output) is determined by whatever number is input for x

For example:

if you input 1 for x, y must then equal 1

If you input 2 for x, y must then equal -2

If you input 3 for x, y must then equal -5

Example # 5
• Rewrite the equation 3x + y = 4 so that x is a function of y
• Use the result to find x when y = -2, -1, and 0

First decide which side of the equation to work on. Since x is on the left begin there

On the left you have a +y. To undo the +y subtract y from both sides of the equation

3x + y = 4

Since 4 and –y are not like terms they cannot be combined

The y’s on the left cancel out leaving 3x

- y -y

3x = 4 – y

3x = 4 – y is NOT in function form. To get the equation in function form x has to be all by itself.

Example #5 (continued)

3x = 4 – y

To get x all by itself you have to undo the multiplication of 3 times x by dividing both sides by 3.

Be careful here!...when you divide both sides by 3 you have to divide the whole right side by 3!

3x = 4 – y

The 3’s on the left cancel out leaving x

3

3

This is the equation in function form with x isolated on the left side of the equation

x = 4 – y

3

x = 4 – y

3

Example #5 (continued)

B. Use the result to find x when y = -2, -1, and 0

Now that you have the equation in function form, simply substitute the value ofy (input) into the equation to determine the value of x (output)

Input

Substitute

Output

x = 4 – (-2)

x = 2

y = -2

3

x = 4 – (-1)

y = -1

x = 5/3

3

x = 4 – 0

x = 4/3

y = 0

3

• Transforming equations and formulas is relatively simple…
• The key to being successful here is being organized, laying out your problem and solving step-by-step.
• This concept will be utilized extensively when we plot linear equations on the coordinate plane later in the course…
• Make no mistake about it…at the level you are required to know how to transform equations and formulas….
• On the next couple of slides are some practice problems…The answers are on the last slide…
• Do the practice and then check your answers…If you do not get the same answer you must question what you did…go back and problem solve to find the error…
• If you cannot find the error bring your work to me and I will help…

Solve for the indicated variable:

• A = ½ bh Solve for b
• V = ℓwh Solve for h
• V = πr2h Solve for h
• A = ½ h(b1 + b2) Solve for b2

Rewrite each equation so that y is a function of x

• 2x + y = 5
• 13 = 12x – 2y
• 9 – y = 1.5x
• y – 7 = -2x

5

• -3x + 4y – 5 = -14
• 1 (25 – 5y) = 4x – 9y + 13

5

b = 2a

h

h = V

ℓw

h = V

πr2

b2 = 2a – b1

h

y = 5 – 2x

y = 6x – 13/2

y = -1.5x + 9

y = -10x + 35

y = ¾ x – 9/4

y = ½ x + 1

Summary
• A key tool in making learning effective is being able to summarize what you learned in a lesson in your own words…
• In this lesson we talked about Formulas and Functions Therefore, in your own words summarize this lesson…be sure to include key concepts that the lesson covered as well as any points that are still not clear to you…
• I will give you credit for doing this lesson…please see the next slide…
Credit
• I will add 25 points as an assignment grade for you working on this lesson…
• To receive the full 25 points you must do the following:
• Have your name, date and period as well a lesson number as a heading.
• Do each of the your turn problems showing all work
• Have a 1 paragraph summary of the lesson in your own words
• Please be advised – I will not give any credit for work submitted: