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Applying Systems of Linear Equations Lesson 6 -5

You solved systems of equations by using substitution and elimination. • Determine the best method for solving systems of equations and apply systems of equations to real-world problems.

Choose the Best Method Determine the best method to solve the system of equations. Then solve the system. 2 x + 3 y = 23 4 x + 2 y = 34 Understand To determine the best method to solve the system of equations, look closely at the coefficients of each term. Plan Since neither the coefficients of x nor the coefficients of y are 1 or – 1, you should not use the substitution method. Since the coefficients are not the same for either x or y, you will need to use elimination with multiplication.

Choose the Best Method Solve Multiply the first equation by – 2 so the coefficients of the x -terms are additive inverses. Then add the equations. 2 x + 3 y = 23 – 4 x – 6 y = – 46 4 x + 2 y = 34 (+) 4 x + 2 y = 34 – 4 y = – 12 y=3 Multiply by – 2. Add the equations. Divide each side by – 4. Simplify.

Choose the Best Method Now substitute 3 for y in either equation to find the value of x. 4 x + 2 y = 34 Second equation 4 x + 2(3) = 34 4 x + 6 – 6 = 34 – 6 4 x = 28 y=3 Simplify. Subtract 6 from each side. Simplify. Divide each side by 4. x =7 Simplify. Answer: The solution is (7, 3).

Choose the Best Method Check Substitute (7, 3) for (x, y) in the first equation. 2 x + 3 y = 23 ? 2(7) + 3(3) = 23 23 = 23 First equation Substitute (7, 3) for (x, y). Simplify.

POOL PARTY At the school party, Mr. Lewis bought 1 adult ticket and 2 child tickets for $10. Mrs. Vroom bought 2 adult tickets and 3 child tickets for $17. The following system can be used to represent this situation, where x is the number of adult tickets and y is the number of child tickets. Determine the best method to solve the system of equations. Then solve the system. x + 2 y = 10 2 x + 3 y = 17 A. substitution; (4, 3) B. substitution; (4, 4) C. elimination; (3, 3) D. elimination; (– 4, – 3)

Apply Systems of Linear Equations CAR RENTAL Ace Car Rental rents a car for $45 and $0. 25 per mile. Star Car Rental rents a car for $35 and $0. 30 per mile. How many miles would a driver need to drive before the cost of renting a car at Ace Car Rental and renting a car at Star Car Rental were the same? Let x = number of miles and y = cost of renting a car. y = 45 + 0. 25 x y = 35 + 0. 30 x

Apply Systems of Linear Equations Subtract the equations to eliminate the y variable. y = 45 + 0. 25 x (–) y = 35 + 0. 30 x 0 = 10 – 0. 05 x Write the equations vertically and subtract. – 10 = – 0. 05 x Subtract 10 from each side. 200 = x Divide each side by – 0. 05.

Apply Systems of Linear Equations Substitute 200 for x in one of the equations. y = 45 + 0. 25 x First equation y = 45 + 0. 25(200) Substitute 200 for x. y = 45 + 50 Simplify. y = 95 Add 45 and 50. Answer: The solution is (200, 95). This means that when the car has been driven 200 miles, the cost of renting a car will be the same ($95) at both rental companies.

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