The Substitution Method

Transcription

1 The Substitution Method As compared to the graphing method, the substitution method of solving a system of linear equations is faster and provides an exact solution. Procedure: To solve a system of linear equations by substitution: 1. Solve one equation for either variable. 2. Substitute the expression equal to the solved variable into the other equation. Solve the resulting equation for the remaining variable. 3. Substitute this value into either equation to solve for the other variable. y = 3 y = 12 Solution: Since the 1 st equation is already solved for y, substitute this expression in for y in the other equation. y = 12 ( 3) = = 12 3x = 15 x = 5 Now solve for y. y = 3 y = 2( 5) 3 y = 13 The ordered pair (-5, -13) is the solution to the system. y = 3x y = 6x 4 Solution: Since both equations are solved for y, set the equations equal to one another. 6x 4 = 3x 3x = 9 x = 3 Now, solve for y using either equation. y = 6x 4 y = 6(3) 4 y 8 4 y 4 The ordered pair (3, 14) is a solution to the system.

2 y 5 x + 3y = 5 Solution: Solve the 2 nd equation for x. This gives the system: y 5 x = 3y Substitute the expression for x into the 1 st equation. y 5 2( 3y ) y 5 6y + 10 y 5 y = 5 y = 5 Solve for x. x = 3y x = 3( 5) x = 20 The ordered pair (20, -5) is the solution to the system. y = 8 = 25 Solution: Solve the first equation for y. y = + 8 = 25 Now substitute the expression for y into the 2nd equation. = 25 5( + 8) = = = 25 This is a contradiction because A contradiction is a statement that is never true. Therefore, there is no solution (inconsistent system.)

3 2( x 3) = 3y 1 y = x Solution: Since the 2 nd equation is solved for y, substitute the expression for y into the 1st equation. 2( x 3) = 3y 1 2( x 3) = 3( x ) 1 6 = 3x x = 20 Solve for y by substituting this value into the 1 st equation. y = x y = 20 y = 15 The ordered pair (-20, -15) is a solution to the system. 3y = y = 24 Solution: Solve the first equation for y to obtain the system. y = 5 x y = 24 Substitute this expression in for y in the 2 nd equation. + 6y = = 24 ( 5 x + 4) + 6 = = 24 This is an identity. An identity is any equation that is always true. Consequently, there are an infinite number of solutions (consistent, dependent system.)

4 4x 3( y + 4) = 2( x 3) + 2y 5 4( y 2) = 5y 1+ 4x Solution: Begin by simplifying each equation to either general form or slope-intercept form to obtain: y = 4x 7 Substitute the expression for y from the 2 nd equation into the 1 st equation. 5( 4x 7) + 20x = 34 x = 17 Substitute this value into the 2 nd equation to solve for y. The ordered pair ( 17, 9 ) ( 17 ) 2 34 = 45 ( 45 )( 1 ) y = y = 9 is a solution to the system. 5 4( p 2) = 4 p q 5q = p 5( q + 4) Solution: Simplify both equations to obtain: q = 15 p = 20 Therefore, the ordered pair (20, -15) is a solution to the system.

5 Applications: Example: Westview Flight School rents its Cessna 172 s for $65.00 per hour and charges $95.00 insurance coverage. Flying Ace Flight School charges $55.00 per hour for the same aircraft and $ for insurance. Create an equation to represent the total cost of renting an aircraft from each flight school, and then find the break-even point. If a pilot wishes to fly for 3 hours, which company should he rent from? Solution: The break-even point is the point where the equations are equal. Westview: C = 65 h + 95 Flying Ace: C = 55 h Since both equations are equal to C, set them equal to each other. 65h + 95 = 55h h = 50 h = 5 Substitute this value into either equation and solve for C. C = 65h + 95 C = 65(5) + 95 C = 420 The break-even point is at h = 5 hours. The cost of flying 5 hours is $ If a pilot wishes to fly 3 hours, which company should he rent from?? To determine this, substitute h=3 into both equations. C = 65h + 95 C = 65(3) + 95 C = 290 C = 55h C = 55(3) C = 310 To fly for 3 hours the pilot should rent from Westview Flight School.

6 Example: Two taxi companies compete in the same neighborhood. One of the companies charges $3.00 for the taxi drop plus $1.25 for each mile driven, while the other charges $2.00 for the taxi drop, plus $1.50 for each mile driven. Create a system of equations, solve and interpret the results. Solution: The two equations are as follows: y y.50x + 2 Since both equations are solved for y, set them equal to each other and solve for x. 1.50x x = 2 Solve for y. y y.25(2) + 3 y = y = 5.5 When we solve this system, we are finding the point using either company will have the same cost (break-even point). This value occurs at x = 2 hours and at a cost of y = $5.50. Example: A laboratory technician needs to make a 10 liter batch of antiseptic that is 60% alcohol. How can he combine a batch of antiseptic that is 30% alcohol with another that is 70% alcohol to get the desired concentration? Solution: Let x = the amount of the 30% alcohol solution and y = the amount of the 70% alcohol solution. We can then create the two equations: x + y 0 0.3x y = 0.6(10) Solve the 1 st equation for x and substitute this expression into the 2 nd equation and solve for y. 0.3(10 y) y = y y = 6 0.4y = 3 y = 7.5 Solve for x. x + y 0 x x = 2.5 It will require 2.5 liters of the 30% solution and 7.5 liters of the 70% solution to create 10 liters of a 60% solution.

7 Example: A $400,000 investment was split so that part was invested at a 7% annual rate of interest and the rest at 9%. If the total earnings were $ how much money was invested at each rate? Solution: Let x = the amount invested at 7% and let y = the amount invested at 9%. We may then create the equations: x + y = 400, x + 0.9y = 3140 Solve the 1st equation for y and substitute this expression into the 2 nd equation. 0.07x (400,000 x) = x + 36, x = = 32,860 x 64,300 Solve for y. y = 400,000 x y = 235,700 $164,300 must be invested at 7$ and $235,700 must be invested at 9%. Example: On a particular airline route, a full price business ticket costs $ and a coach ticket costs $ On one of these flights, there were 172 passengers which resulted in a total ticket income of $44, How many business tickets were sold? Solution: Let b = the number of business tickets and c = the number of coach tickets sold. We may then write the following system of equations: b + c b + 210c = 44,120 Solve the 1 st equation for c and substitute this expression into the 2 nd equation and solve for b. 310b + 210(172 b) = 44, b b = 44, b = 8,000 b = 80 Therefore 80 business tickets were sold.