Solving Rational Equations Section 7.5 beginning on page 392 Solving by Cross Multiplying When you have two equal ratios, their cross products are equal. Example 1: Solve 3 9 = π₯ + 1 4π₯ + 5 3(4π₯ + 5) = 9(π₯ + 1) 12π₯ + 15 = 9π₯ + 9 3π₯ + 15 = 9 3π₯ = −6 π₯ = −2 CHECK: 3 9 = (−2) + 1 4(−2) + 5 3 9 = −1 −8 + 5 −3 = 9 −3 −3 = −3 Solving by Using the LCD Example 3a : Solve 4π₯ 5 7 4π₯ 9 4π₯ β + =− β β 1 π₯ 4 1 π₯ 1 ** Multiply each term by the LCD. πΏπΆπ· = 4π₯ ** Check 20π₯ 28π₯ 36π₯ + =− π₯ 4 π₯ 20 + 7π₯ = −36 7π₯ = −56 π₯ = −8 ** Simplify ** Solve 5 7 2 9 + β =− (−8) 4 2 (−8) 5 14 9 − + = 8 8 8 −5 + 14 9 = 8 8 9 9 = 8 8 Solving by Using the LCD Example 3b: Solve 1− 8 3 = π₯−5 π₯ π₯(π₯ − 5) β 1 − πΏπΆπ· = π₯(π₯ − 5) 8 π₯(π₯ − 5) 3 π₯(π₯ − 5) β = β π₯−5 1 π₯ 1 π₯(π₯ − 5) − 8π₯(π₯ − 5) 3π₯(π₯ − 5) = π₯−5 π₯ π₯ 2 − 5π₯ − 8π₯ = 3(π₯ − 5) π₯ 2 − 13π₯ = 3π₯ − 15 π₯ 2 − 16π₯ + 15 = 0 π₯ − 15 π₯ − 1 = 0 π₯ = 15, π₯ = 1 Solving an Equation With an Extraneous Solution Example 4: Solve 6 π₯−3 6 8π₯ 2 4π₯ = − π₯ − 3 π₯2 − 9 π₯ + 3 8π₯ 2 4π₯ π₯ + 3 (π₯ − 3) = π₯ + 3 (π₯ − 3) − π₯ + 3 (π₯ − 3) π₯+3 6 π₯ + 3 = 8π₯ 2 − 4π₯(π₯ − 3) 6π₯ + 18 = 8π₯ 2 − 4π₯ 2 + 12π₯ 6π₯ + 18 = 4π₯ 2 + 12π₯ 0 = 4π₯ 2 + 6π₯ − 18 0 = 2π₯ 2 + 3π₯ − 9 0 = (2π₯ − 3)(π₯ + 3) 3 π₯= , 2 π₯ = −3 π₯ + 3 (π₯ − 3) Solving a Real – Life Problem 50π + 1000 Example 6 : The function π = represents the average cost c (in π dollars) of making m models using a 3-D printer. Find how many models must be printed for the average cost per model to fall to $90. π β 90 = 50π + 1000 βπ π 90π = 50π + 1000 40π = 1000 π = 25 The average cost per model falls to $90 after 25 have been printed. Monitoring Progress Solve each equation and check your solution(s). 1) 3 2 = 5π₯ π₯ − 7 3) 1 π₯ = 2π₯ + 5 11π₯ + 8 5) 3π₯ 5 3 − = π₯ + 1 2π₯ 2π₯ 2 π₯ = − ,π₯ = 2 3 π₯ = −3 π₯ = −1, π₯=4 1 3 2) −4 5 = π₯+3 π₯−3 π₯=− 4) 15 4 7 + = π₯ 5 π₯ π₯ = −10 6) 4π₯ + 1 12 = 2 +3 π₯+1 π₯ −1 π₯ = −2, π₯ = 5 Monitoring Progress Solve each equation and check your solution(s). 7) 9 6π₯ 9π₯ 2 + = π₯ − 2 π₯ + 2 π₯2 − 4 8) 7 6 −5= 2 π₯−1 π₯ −1 π₯ = −3 3 π₯ = − ,π₯ = 2 5 1 **9) Consider the function π π₯ = − 2 . Determine whether the inverse of π is a π₯ function then find the inverse. π¦ππ , π‘βπ πππ£ππ π ππ π ππ’πππ‘πππ. π π₯ = 1 π₯+2