7.5: Solving Rational Equations

advertisement
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‘₯
βˆ™
+
=− βˆ™
βˆ™
1 π‘₯
4 1
π‘₯ 1
** Multiply each term by the LCD.
𝐿𝐢𝐷 = 4π‘₯
** Check
20π‘₯ 28π‘₯
36π‘₯
+
=−
π‘₯
4
π‘₯
20 + 7π‘₯ = −36
7π‘₯ = −56
π‘₯ = −8
** Simplify
** Solve
5
7 2
9
+ βˆ™ =−
(−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) βˆ™ 1 −
𝐿𝐢𝐷 = π‘₯(π‘₯ − 5)
8
π‘₯(π‘₯ − 5) 3 π‘₯(π‘₯ − 5)
βˆ™
= βˆ™
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‘š + 1000
Example 6 : The function 𝑐 =
represents the average cost c (in
π‘š
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‘š βˆ™ 90 =
50π‘š + 1000
βˆ™π‘š
π‘š
90π‘š = 50π‘š + 1000
40π‘š = 1000
π‘š = 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 𝑓 π‘₯ = − 2 . Determine whether the inverse of 𝑓 is a
π‘₯
function then find the inverse.
𝑦𝑒𝑠, π‘‘β„Žπ‘’ 𝑖𝑛𝑣𝑒𝑠𝑒 𝑖𝑠 π‘Ž π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘›.
𝑔 π‘₯ =
1
π‘₯+2
Download