Chapter 2.6 Formulas & Functions Algebra 1 Chapter 2.6 Formulas & Functions

Objective Students will solve formula’s for one of its variables and rewrite equations in function form. 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…

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

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

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

Summary 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…

Comments On the next slide are some practice problems… 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…

Homework 2.6 Solve for the indicated variable: Solve for b Solve for h