1. Solve Linear Systems by Substitution Section 6.2 beginning on page 337.

2. Solving by Substitution This method is important to learn because it will be the best method when having to solve a system with more than 2 variables. To solve a system using substitution: Solve one equation for a variable (make a smart choice) Substitute into the other equation and solve it. Substitute your known value into either equation to find the second value. Display your answer as a coordinate pair.

3. Substitution Examples When a variable is already alone in one equation, use that to substitute. Don’t do more work than you have to! *When no variables have a coefficient of 1, try to find a variable with a coefficient that is a factor of each term in the equation. 1) (-1,6) 2) (2,-1) 3) (2,-1) **You can check your solution by substituting into each of the equations. ** (this is also how you would “check to see if the given point is a solution to the system”)

4. Word Problem - Comparison Example 3 (pg. 379): Many businesses pay website hosting companies to store and maintain the computer files that make up their websites. Internet service providers also offer website hosting. The costs for website hosting offered by a website hosting company and an Internet service provider are shown in the table. Find the number of months after which to total cost for website hosting will be the same for both companies. Hint: Each equation will represent how the total cost is calculated for each company. The solution to the system will represent the number of months where each company costs the same, and how much that cost is. CompanySet-up fee (dollars)Cost per month (dollars) Internet Service Provider1021.95 Website Hosting CompanyNone22.45

5. Solution CompanySet-up fee (dollars)Cost per month (dollars) Internet Service Provider1021.95 Website Hosting CompanyNone22.45 First “declare” your variables !!y = The total cost, x = number of months The total cost will be the same for the two companies at 20 months, it will be $449 for both.

6. Word Problem - Mixture Example 4 (page 380): For extremely cold temperatures, an automobile manufacturer recommends that a 70% antifreeze and 30% water mix be used in the cooling system of a car. How many quarts of pure (100%) antifreeze and a 50% antifreeze and 50% water mix should be combined to make 11 quarts of a 70% antifreeze and 30% water mix? Hint: One equation will be a calculation of how much “stuff” there is (in this specific situation it is quarts of liquid), and the other equation calculates the parts or percent's of the mixtures creating the final mixture (in this case, the amount of antifreeze in the solution). Declare your variables! Y= quarts of pure antifreeze X= quarts of 50-50 mix Write known information: We are making 11 quarts total The mix will be 70% antifreeze and 30% water.

7. Solution Y= quarts of pure antifreeze X= quarts of 50-50 mix We are making 11 quarts total Quarts of pure + Quarts of 50% antifreeze = Quarts of final mixture The mix will be 70% antifreeze and 30% water. 100% of the pure mix + 50% of the 50/50 mix = 70% of the final mix 6.6 quarts of the 50-50 mix and 4.4 quarts of the pure antifreeze.